金沙国际:项目涉嫌到的肆14个Sql语句

sql server 查询记录平均值及并列排在一条线序
的话语查询学子的平分成绩并张开排名,sql
2003用子查询达成,分平均战绩再度时保留排名空缺和不保留排行空缺三种。select
t1.* , px = (select count(1) from ( select m.S# [学员编号] , m.Sname
[学员姓名] , isnull(cast(avg(score) as decimal(18,2)),0) [平均成绩]
from Student m left join SC n on m.S# = n.S# group by m.S# , m.Sname卡塔尔(قطر‎t2 where 平均成绩 t1.平均成绩卡塔尔(قطر‎ + 1 from ( select m.S# [学子编号] ,
m.Sname [学子姓名] , isnull(cast(avg(score) as decimal(18,2)),0)
[平均战绩] from Student m left join SC n on m.S# = n.S# group by
m.S# , m.Sname) t1order by px

青云

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/*
FROM CSDN
表明:以下47个语句都听从测量检验数据开展过测验,最棒每一回只单独运营八个说话。
标题及描述:
–1.学生表
Student(S#,Sname,Sage,Ssex) –S# 学子编号,Sname 学子姓名,Sage
出生年月,Ssex 学子性别
–2.课程表
Course(C#,Cname,T#) –C# –课程编号,Cname 课程名称,T# 教授编号
–3.教师表
Teacher(T#,Tname) –T# 教授编号,Tname 教授姓名
–4.成绩表
SC(S#,C#,score) –S# 学子编号,C# 课程编号,score 分数
*/
–成立测验数据
create table Student(S# varchar(10),Sname nvarchar(10),Sage datetime,Ssex nvarchar(10))
insert into Student values(’01’ , N’赵雷’ , ‘1990-01-01′ , N’男’)
insert into Student values(’02’ , N’钱电’ , ‘1990-12-21′ , N’男’)
insert into Student values(’03’ , N’孙风’ , ‘1990-05-20′ , N’男’)
insert into Student values(’04’ , N’李云’ , ‘1990-08-06′ , N’男’)
insert into Student values(’05’ , N’周梅’ , ‘1991-12-01′ , N’女’)
insert into Student values(’06’ , N’吴兰’ , ‘1992-03-01′ , N’女’)
insert into Student values(’07’ , N’郑竹’ , ‘1989-07-01′ , N’女’)
insert into Student values(’08’ , N’王菊’ , ‘1990-01-20′ , N’女’)
create table Course(C# varchar(10),Cname nvarchar(10),T# varchar(10))
insert into Course values(’01’ , N’语文’ , ’02’)
insert into Course values(’02’ , N’数学’ , ’01’)
insert into Course values(’03’ , N’英语’ , ’03’)
create table Teacher(T# varchar(10),Tname nvarchar(10))
insert into Teacher values(’01’ , N’张三’)
insert into Teacher values(’02’ , N’李四’)
insert into Teacher values(’03’ , N’王五’)
create table SC(S# varchar(10),C# varchar(10),score decimal(18,1))
insert into SC values(’01’ , ’01’ , 80)
insert into SC values(’01’ , ’02’ , 90)
insert into SC values(’01’ , ’03’ , 99)
insert into SC values(’02’ , ’01’ , 70)
insert into SC values(’02’ , ’02’ , 60)
insert into SC values(’02’ , ’03’ , 80)
insert into SC values(’03’ , ’01’ , 80)
insert into SC values(’03’ , ’02’ , 80)
insert into SC values(’03’ , ’03’ , 80)
insert into SC values(’04’ , ’01’ , 50)
insert into SC values(’04’ , ’02’ , 30)
insert into SC values(’04’ , ’03’ , 20)
insert into SC values(’05’ , ’01’ , 76)
insert into SC values(’05’ , ’02’ , 87)
insert into SC values(’06’ , ’01’ , 31)
insert into SC values(’06’ , ’03’ , 34)
insert into SC values(’07’ , ’02’ , 89)
insert into SC values(’07’ , ’03’ , 98)
go

select t1.* , px = (select count(distinct 平均成绩State of Qatar from ( select m.S#
[学子编号] , m.Sname [学子姓名] , isnull(cast(avg(score) as
decimal(18,2)),0) [平均战绩] from Student m left join SC n on m.S# =
n.S# group by m.S# , m.Sname卡塔尔国 t2 where 平均成绩 = t1.平均成绩卡塔尔国 from (
select m.S# [学员编号] , m.Sname [学员姓名] ,
isnull(cast(avg(score) as decimal(18,2)),0) [平均成绩] from Student m
left join SC n on m.S# = n.S# group by m.S# , m.Sname) t1order by px

三个项目涉及到的五19个Sql语句(收拾版卡塔尔(قطر‎

/*标题:一个项目涉及到的50个Sql语句(整理版)说明:以下五十个语句都按照测试数据进行过测试,最好每次只单独运行一个语句。*/
--1.学生表
Student(S,Sname,Sage,Ssex) --S 学生编号,Sname 学生姓名,Sage 出生年月,Ssex 学生性别--2.课程表 
Course(C,Cname,T) --C --课程编号,Cname 课程名称,T 教师编号
--3.教师表 
Teacher(T,Tname) --T 教师编号,Tname 教师姓名--4.成绩表 
SC(S,C,score) --S 学生编号,C 课程编号,score 分数*/--创建测试数据
create table Student(S varchar(10),Sname nvarchar(10),Sage datetime,Ssex nvarchar(10))insert into Student values('01' , N'赵雷' , '1990-01-01' , N'男')insert into Student values('02' , N'钱电' , '1990-12-21' , N'男')insert into Student values('03' , N'孙风' , '1990-05-20' , N'男')insert into Student values('04' , N'李云' , '1990-08-06' , N'男')insert into Student values('05' , N'周梅' , '1991-12-01' , N'女')insert into Student values('06' , N'吴兰' , '1992-03-01' , N'女')insert into Student values('07' , N'郑竹' , '1989-07-01' , N'女')insert into Student values('08' , N'王菊' , '1990-01-20' , N'女')create table Course(C varchar(10),Cname nvarchar(10),T varchar(10))insert into Course values('01' , N'语文' , '02')insert into Course values('02' , N'数学' , '01')insert into Course values('03' , N'英语' , '03')create table Teacher(T varchar(10),Tname nvarchar(10))insert into Teacher values('01' , N'张三')insert into Teacher values('02' , N'李四')insert into Teacher values('03' , N'王五')create table SC(S varchar(10),C varchar(10),score decimal(18,1))insert into SC values('01' , '01' , 80)
insert into SC values('01' , '02' , 90)insert into SC values('01' , '03' , 99)
insert into SC values('02' , '01' , 70)insert into SC values('02' , '02' , 60)
insert into SC values('02' , '03' , 80)insert into SC values('03' , '01' , 80)
insert into SC values('03' , '02' , 80)insert into SC values('03' , '03' , 80)
insert into SC values('04' , '01' , 50)insert into SC values('04' , '02' , 30)
insert into SC values('04' , '03' , 20)insert into SC values('05' , '01' , 76)
insert into SC values('05' , '02' , 87)insert into SC values('06' , '01' , 31)
insert into SC values('06' , '03' , 34)insert into SC values('07' , '02' , 89)
insert into SC values('07' , '03' , 98)go

–1、查询”01″课程比”02″课程战表高的学员的音讯及学科分数
–1.1、查询同一时间设有”01″课程和”02″课程的动静
select a.* , b.score [课程’01’的分数],c.score [课程’02’的分数] from
Student a , SC b , SC c 
where a.S = b.S and a.S = c.S and b.C = ’01’ and c.C = ’02’ and b.score
> c.score
–1.2、查询同不经常间存在”01″课程和”02″课程的景况和存在”01″课程但或者荒诞不经”02″课程的意况(海市蜃楼时显得为null卡塔尔(قطر‎(以下存在同样内容时不再解释卡塔尔国
select a.* , b.score [课程”01″的分数],c.score [课程”02″的分数] from
Student a 
left join SC b on a.S = b.S and b.C = ’01’
left join SC c on a.S = c.S and c.C = ’02’
where b.score > isnull(c.score,0)

–2、查询”01″课程比”02″课程战绩低的上学的小孩子的消息及教程分数
–2.1、查询同时存在”01″课程和”02″课程的景观
select a.* , b.score [课程’01’的分数],c.score [课程’02’的分数] from
Student a , SC b , SC c 
where a.S = b.S and a.S = c.S and b.C = ’01’ and c.C = ’02’ and b.score
< c.score
–2.2、查询同一时间存在”01″课程和”02″课程的景况和不设有”01″课程但存在”02″课程的事态
select a.* , b.score [课程”01″的分数],c.score [课程”02″的分数] from
Student a 
left join SC b on a.S = b.S and b.C = ’01’
left join SC c on a.S = c.S and c.C = ’02’
where isnull(b.score,0) < c.score

–3、查询平均成绩超过等于60分的同窗的学子编号和学习者姓名和平均成绩
select a.S , a.Sname , cast(avg(b.score) as decimal(18,2)) avg_score
from Student a , sc b
where a.S = b.S
group by a.S , a.Sname
having cast(avg(b.score) as decimal(18,2)) >= 60 
order by a.S

–4、查询平均成绩小于60分的同校的学员编号和学员姓名和平均战表
–4.1、查询在sc表存在战表的学员音讯的SQL语句。
select a.S , a.Sname , cast(avg(b.score) as decimal(18,2)) avg_score
from Student a , sc b
where a.S = b.S
group by a.S , a.Sname
having cast(avg(b.score) as decimal(18,2)) < 60 
order by a.S
–4.2、查询在sc表中子虚乌有战绩的学子新闻的SQL语句。
select a.S , a.Sname , isnull(cast(avg(b.score) as decimal(18,2)),0)
avg_score
from Student a left join sc b
on a.S = b.S
group by a.S , a.Sname
having isnull(cast(avg(b.score) as decimal(18,2)),0) < 60 
order by a.S

–5、查询全部同学的学子编号、学子姓名、选课总的数量、全数课程的总成绩
–5.1、查询全数有成绩的SQL。
select a.S [学子编号], a.Sname [学生姓名], count(b.CState of Qatar 选课总量,
sum(score卡塔尔 [具备科目标总成绩]
from Student a , SC b 
where a.S = b.S 
group by a.S,a.Sname 
order by a.S
–5.2、查询全体(包括有成就和无战绩卡塔尔(قطر‎的SQL。
select a.S [学生编号], a.Sname [学子姓名], count(b.C卡塔尔(قطر‎ 选课总的数量,
sum(score卡塔尔 [全部科目标总战绩]
from Student a left join SC b 
on a.S = b.S 
group by a.S,a.Sname 
order by a.S

–6、查询”李”姓老师的数码 
–方法1
select count(Tname) [“李”姓老师的多少] from Teacher where Tname like
N’李%’
–方法2
select count(Tname) [“李”姓老师的数量] from Teacher where
left(Tname,1) = N’李’
/*
“李”姓老师的数码   
———– 
1
*/

–7、查询学过”张三”老师讲课的同校的消息 
select distinct Student.* from Student , SC , Course , Teacher 
where Student.S = SC.S and SC.C = Course.C and Course.T = Teacher.T and
Teacher.Tname = N’张三’
order by Student.S

–8、查询没学过”张三”老师教学的同窗的音讯 
select m.* from Student m where S not in (select distinct SC.S from SC
, Course , Teacher where SC.C = Course.C and Course.T = Teacher.T and
Teacher.Tname = N’张三’) order by m.S

–9、查询学过数码为”01″並且也学过数码为”02″的学科的同校的音讯
–方法1
select Student.* from Student , SC where Student.S = SC.S and SC.C =
’01’ and exists (Select 1 from SC SC_2 where SC_2.S = SC.S and SC_2.C
= ’02’) order by Student.S
–方法2
select Student.* from Student , SC where Student.S = SC.S and SC.C =
’02’ and exists (Select 1 from SC SC_2 where SC_2.S = SC.S and SC_2.C
= ’01’) order by Student.S
–方法3
select m.* from Student m where S in
(
  select S from
  (
    select distinct S from SC where C = ’01’
    union all
    select distinct S from SC where C = ’02’
  ) t group by S having count(1) = 2 
)
order by m.S

–10、查询学过数码为”01″不过未有学过数码为”02″的科指标校友的消息
–方法1
select Student.* from Student , SC where Student.S = SC.S and SC.C =
’01’ and not exists (Select 1 from SC SC_2 where SC_2.S = SC.S and
SC_2.C = ’02’) order by Student.S
–方法2
select Student.* from Student , SC where Student.S = SC.S and SC.C =
’01’ and Student.S not in (Select SC_2.S from SC SC_2 where SC_2.S =
SC.S and SC_2.C = ’02’) order by Student.S

–11、查询未有学全全数科指标同室的音讯 
–11.1、
select Student.*
from Student , SC 
where Student.S = SC.S 
group by Student.S , Student.Sname , Student.Sage , Student.Ssex having
count(C) < (select count(C) from Course) 
–11.2
select Student.*
from Student left join SC 
on Student.S = SC.S 
group by Student.S , Student.Sname , Student.Sage , Student.Ssex having
count(C) < (select count(C) from Course)

–12、查询至罕有一门课与学号为”01″的校友所学形似的校友的音讯 
select distinct Student.* from Student , SC where Student.S = SC.S and
SC.C in (select C from SC where S = ’01’) and Student.S <> ’01’

–13、查询和”01″号的同学学习的学科完全肖似的其余同学的消息 
select Student.* from Student where S in
(select distinct SC.S from SC where S <> ’01’ and SC.C in (select
distinct C from SC where S = ’01’) 
group by SC.S having count(1) = (select count(1) from SC where S=’01’))

–14、查询没学过”张三”老师教授的任一门科目的学子姓名 
select student.* from student where student.S not in 
(select distinct sc.S from sc , course , teacher where sc.C = course.C
and course.T = teacher.T and teacher.tname = N’张三’)
order by student.S

–15、查询两门及其以上不如格课程的同窗的学号,姓名及其平均战绩 
select student.S , student.sname , cast(avg(score) as decimal(18,2))
avg_score from student , sc 
where student.S = SC.S and student.S in (select S from SC where score
< 60 group by S having count(1) >= 2)
group by student.S , student.sname

–16、检索”01″课程分数小于60,按分数降序排列的学生信息
select student.* , sc.C , sc.score from student , sc 
where student.S = SC.S and sc.score < 60 and sc.C = ’01’
order by sc.score desc 

–17、按平均成绩从高到低展现全体学子的享有课程的大成以至平均战绩
–17.1 SQL 2000 静态 
select a.S 学子编号 , a.Sname 学子姓名 ,
       max(case c.Cname when N’语文’ then b.score else null end)
[语文],
       max(case c.Cname when N’数学’ then b.score else null end)
[数学],
       max(case c.Cname when N’英语’ then b.score else null end)
[英语],
       cast(avg(b.score) as decimal(18,2)) 平均分
from Student a 
left join SC b on a.S = b.S
left join Course c on b.C = c.C
group by a.S , a.Sname
order by 平均分 desc
–17.2 SQL 2000 动态 
declare @sql nvarchar(4000)
set @sql = ‘select a.S ‘ + N’学子编号’ + ‘ , a.Sname ‘ + N’学子姓名’
select @sql = @sql + ‘,max(case c.Cname when N”’+Cname+”’ then b.score
else null end) [‘+Cname+’]’
from (select distinct Cname from Course) as t
set @sql = @sql + ‘ , cast(avg(b.score) as decimal(18,2)) ‘ + N’平均分’

  • ‘ from Student a left join SC b on a.S = b.S left join Course c on b.C
    = c.C
    group by a.S , a.Sname order by ‘ + N’平均分’ + ‘ desc’
    exec(@sql)
    –17.3 有关sql 二〇〇七的动静态写法参见作者的篇章《普通行列调换(version
    2.0卡塔尔(قطر‎》或《普通行列调换(version 3.0State of Qatar》。

 

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#1楼 得分:0回复于:2010-05-17 17:47:07
SQL code
–18、查询各科战绩最高分、最低分和平均分:以如下情势显得:课程ID,课程name,最高分,最低分,平均分,及格率,中等率,杰出率,优质率
–及格为>=60,中等为:70-80,优良为:80-90,优秀为:>=90
–方法1
select m.C [学科编号], m.Cname [课程名称], 
  max(n.score) [最高分],
  min(n.score) [最低分],
  cast(avg(n.score) as decimal(18,2)) [平均分],
  cast((select count(1) from SC where C = m.C and score >= 60)*100.0
/ (select count(1) from SC where C = m.C) as decimal(18,2))
[及格率(%)],
  cast((select count(1) from SC where C = m.C and score >= 70 and
score < 80 )*100.0 / (select count(1) from SC where C = m.C) as
decimal(18,2)) [中等率(%)],
  cast((select count(1) from SC where C = m.C and score >= 80 and
score < 90 )*100.0 / (select count(1) from SC where C = m.C) as
decimal(18,2)) [优良率(%)],
  cast((select count(1) from SC where C = m.C and score >= 90)*100.0
/ (select count(1) from SC where C = m.C) as decimal(18,2))
[优秀率(%)]
from Course m , SC n
where m.C = n.C
group by m.C , m.Cname
order by m.C
–方法2
select m.C [课程编号], m.Cname [课程名称], 
  (select max(score) from SC where C = m.C) [最高分],
  (select min(score) from SC where C = m.C) [最低分],
  (select cast(avg(score) as decimal(18,2)) from SC where C = m.C)
[平均分],
  cast((select count(1) from SC where C = m.C and score >= 60)*100.0
/ (select count(1) from SC where C = m.C) as decimal(18,2))
[及格率(%)],
  cast((select count(1) from SC where C = m.C and score >= 70 and
score < 80 )*100.0 / (select count(1) from SC where C = m.C) as
decimal(18,2)) [中等率(%)],
  cast((select count(1) from SC where C = m.C and score >= 80 and
score < 90 )*100.0 / (select count(1) from SC where C = m.C) as
decimal(18,2)) [优良率(%)],
  cast((select count(1) from SC where C = m.C and score >= 90)*100.0
/ (select count(1) from SC where C = m.C) as decimal(18,2))
[优秀率(%)]
from Course m 
order by m.C

–19、按各科战绩实行排序,并展现排行
–19.1 sql 二〇〇一用子查询完结
–Score重复时保留排名空缺
select t.* , px = (select count(1) from SC where C = t.C and score >
t.score) + 1 from sc t order by t.C , px 
–Score重复时归并排行
select t.* , px = (select count(distinct score) from SC where C = t.C
and score >= t.score) from sc t order by t.C , px 
–19.2 sql 2005用rank,DENSE_RANK完成
–Score重复时保留名次空缺(rank实现State of Qatar
select t.* , px = rank() over(partition by C order by score desc) from
sc t order by t.C , px 
–Score重复时归总排行(DENSE_RANK完成)
select t.* , px = DENSE_RANK() over(partition by C order by score
desc) from sc t order by t.C , px

–20、查询学子的总成绩并开展排名
–20.1 查询学子的总战表
select m.S [学员编号] , 
       m.Sname [学员姓名] ,
       isnull(sum(score),0) [总成绩]
from Student m left join SC n on m.S = n.S 
group by m.S , m.Sname
order by [总成绩] desc
–20.2 查询学子的总战表并打开排行,sql
二零零二用子查询完毕,分总分重复时保留排名空缺和不保留排行空缺三种。
select t1.* , px = (select count(1) from 
(
  select m.S [学员编号] , 
         m.Sname [学子姓名] ,
         isnull(sum(score),0) [总成绩]
  from Student m left join SC n on m.S = n.S 
  group by m.S , m.Sname
) t2 where 总成绩 > t1.总成绩) + 1 from 
(
  select m.S [学子编号] , 
         m.Sname [学员姓名] ,
         isnull(sum(score),0) [总成绩]
  from Student m left join SC n on m.S = n.S 
  group by m.S , m.Sname
) t1
order by px

select t1.* , px = (select count(distinct 总成绩) from 
(
  select m.S [学员编号] , 
         m.Sname [学员姓名] ,
         isnull(sum(score),0) [总成绩]
  from Student m left join SC n on m.S = n.S 
  group by m.S , m.Sname
) t2 where 总成绩 >= t1.总成绩) from 
(
  select m.S [学子编号] , 
         m.Sname [学子姓名] ,
         isnull(sum(score),0) [总成绩]
  from Student m left join SC n on m.S = n.S 
  group by m.S , m.Sname
) t1
order by px
–20.3 查询学子的总成绩并开展排行,sql
2006用rank,DENSE_RANK完结,分总分重复时保留排行空缺和不保留排行空缺二种。
select t.* , px = rank() over(order by [总成绩] desc) from
(
  select m.S [学员编号] , 
         m.Sname [学员姓名] ,
         isnull(sum(score),0) [总成绩]
  from Student m left join SC n on m.S = n.S 
  group by m.S , m.Sname
) t
order by px

select t.* , px = DENSE_RANK() over(order by [总成绩] desc) from
(
  select m.S [学员编号] , 
         m.Sname [学子姓名] ,
         isnull(sum(score),0) [总成绩]
  from Student m left join SC n on m.S = n.S 
  group by m.S , m.Sname
) t
order by px

–21、查询不相同老师所教分裂科目平均分从高到低展现 
select m.T , m.Tname , cast(avg(o.score) as decimal(18,2)) avg_score
from Teacher m , Course n , SC o
where m.T = n.T and n.C = o.C
group by m.T , m.Tname
order by avg_score desc

–22、查询全体课程的实际业绩第2名到第3名的学子音讯及该课程战绩
–22.1 sql 二〇〇〇用子查询完毕
–Score重复时保留排名空缺
select * from (select t.* , px = (select count(1) from SC where C =
t.C and score > t.score) + 1 from sc t) m where px between 2 and 3
order by m.C , m.px 
–Score重复时归并排行
select * from (select t.* , px = (select count(distinct score) from SC
where C = t.C and score >= t.score) from sc t) m where px between 2
and 3 order by m.C , m.px 
–22.2 sql 2005用rank,DENSE_RANK完成
–Score重复时保留排名空缺(rank完毕卡塔尔国
select * from (select t.* , px = rank() over(partition by C order by
score desc) from sc t) m where px between 2 and 3 order by m.C , m.px 
–Score重复时归并排名(DENSE_RANK完成)
select * from (select t.* , px = DENSE_RANK() over(partition by C
order by score desc) from sc t) m where px between 2 and 3 order by m.C
, m.px

–23、总结各科战绩各分数段人数:课程编号,课程名称,[100-85],[85-70],[70-60],[0-60]及所占百分比 
–23.1
总结各科成绩各分数段人数:课程编号,课程名称,[100-85],[85-70],[70-60],[0-60]
–横向呈现
select Course.C [课程编号] , Cname as [课程名称] ,
  sum(case when score >= 85 then 1 else 0 end) [85-100],
  sum(case when score >= 70 and score < 85 then 1 else 0 end)
[70-85],
  sum(case when score >= 60 and score < 70 then 1 else 0 end)
[60-70],
  sum(case when score < 60 then 1 else 0 end) [0-60]
from sc , Course 
where SC.C = Course.C 
group by Course.C , Course.Cname
order by Course.C
–纵向呈现1(展现存在的分数段卡塔尔国
select m.C [课程编号] , m.Cname [课程名称] , 分数段 = (
  case when n.score >= 85 then ’85-100′
       when n.score >= 70 and n.score < 85 then ’70-85′
       when n.score >= 60 and n.score < 70 then ’60-70′
       else ‘0-60’
  end) , 
  count(1) 数量 
from Course m , sc n
where m.C = n.C 
group by m.C , m.Cname , (
  case when n.score >= 85 then ’85-100′
       when n.score >= 70 and n.score < 85 then ’70-85′
       when n.score >= 60 and n.score < 70 then ’60-70′
       else ‘0-60’
  end)
order by m.C , m.Cname , 分数段
–纵向呈现2(显示存在的分数段,不设有的分数段用0彰显卡塔尔
select m.C [学科编号] , m.Cname [课程名称] , 分数段 = (
  case when n.score >= 85 then ’85-100′
       when n.score >= 70 and n.score < 85 then ’70-85′
       when n.score >= 60 and n.score < 70 then ’60-70′
       else ‘0-60’
  end) , 
  count(1) 数量 
from Course m , sc n
where m.C = n.C 
group by all m.C , m.Cname , (
  case when n.score >= 85 then ’85-100′
       when n.score >= 70 and n.score < 85 then ’70-85′
       when n.score >= 60 and n.score < 70 then ’60-70′
       else ‘0-60’
  end)
order by m.C , m.Cname , 分数段

–23.2
计算各科战绩各分数段人数:课程编号,课程名称,[100-85],[85-70],[70-60],[<60]及所占百分比 
–横向展现
select m.C 课程编号, m.Cname 课程名称,
  (select count(1) from SC where C = m.C and score < 60) [0-60],
  cast((select count(1) from SC where C = m.C and score < 60)*100.0
/ (select count(1) from SC where C = m.C) as decimal(18,2))
[百分比(%)],
  (select count(1) from SC where C = m.C and score >= 60 and score
< 70) [60-70],
  cast((select count(1) from SC where C = m.C and score >= 60 and
score < 70)*100.0 / (select count(1) from SC where C = m.C) as
decimal(18,2)) [百分比(%)],
  (select count(1) from SC where C = m.C and score >= 70 and score
< 85) [70-85],
  cast((select count(1) from SC where C = m.C and score >= 70 and
score < 85)*100.0 / (select count(1) from SC where C = m.C) as
decimal(18,2)) [百分比(%)],
  (select count(1) from SC where C = m.C and score >= 85)
[85-100],
  cast((select count(1) from SC where C = m.C and score >= 85)*100.0
/ (select count(1) from SC where C = m.C) as decimal(18,2))
[百分比(%)]
from Course m 
order by m.C
–纵向展现1(呈现存在的分数段State of Qatar
select m.C [学科编号] , m.Cname [课程名称] , 分数段 = (
  case when n.score >= 85 then ’85-100′
       when n.score >= 70 and n.score < 85 then ’70-85′
       when n.score >= 60 and n.score < 70 then ’60-70′
       else ‘0-60’
  end) , 
  count(1) 数量 ,  
  cast(count(1) * 100.0 / (select count(1) from sc where C = m.C) as
decimal(18,2)) [百分比(%)]
from Course m , sc n
where m.C = n.C 
group by m.C , m.Cname , (
  case when n.score >= 85 then ’85-100′
       when n.score >= 70 and n.score < 85 then ’70-85′
       when n.score >= 60 and n.score < 70 then ’60-70′
       else ‘0-60’
  end)
order by m.C , m.Cname , 分数段
–纵向呈现2(显示存在的分数段,空头支票的分数段用0显示State of Qatar
select m.C [课程编号] , m.Cname [课程名称] , 分数段 = (
  case when n.score >= 85 then ’85-100′
       when n.score >= 70 and n.score < 85 then ’70-85′
       when n.score >= 60 and n.score < 70 then ’60-70′
       else ‘0-60’
  end) , 
  count(1) 数量 ,  
  cast(count(1) * 100.0 / (select count(1) from sc where C = m.C) as
decimal(18,2)) [百分比(%)]
from Course m , sc n
where m.C = n.C 
group by all m.C , m.Cname , (
  case when n.score >= 85 then ’85-100′
       when n.score >= 70 and n.score < 85 then ’70-85′
       when n.score >= 60 and n.score < 70 then ’60-70′
       else ‘0-60’
  end)
order by m.C , m.Cname , 分数段

 

对本人有用[6]丢个板砖[0]援用举报管理TOP
精粹推荐:SQL语句优化汇总

dawugui
(爱新觉罗.毓华卡塔尔(قطر‎
等 级:
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越来越多勋章
#2楼 得分:0回复于:2010-05-17 17:47:22
SQL code
–24、查询学一生均成绩及其排行 
–24.1 查询学子的平均战绩并拓宽排行,sql
二零零零用子查询完结,分平均战表又一次时保留排名空缺和不保留排名空缺二种。
select t1.* , px = (select count(1) from 
(
  select m.S [学子编号] , 
         m.Sname [学子姓名] ,
         isnull(cast(avg(score) as decimal(18,2)),0) [平均战绩]
  from Student m left join SC n on m.S = n.S 
  group by m.S , m.Sname
State of Qatar t2 where 平均成绩 > t1.平分战绩State of Qatar + 1 from 
(
  select m.S [学员编号] , 
         m.Sname [学员姓名] ,
         isnull(cast(avg(score) as decimal(18,2)),0) [平均成绩]
  from Student m left join SC n on m.S = n.S 
  group by m.S , m.Sname
) t1
order by px

select t1.* , px = (select count(distinct 平均战绩卡塔尔(قطر‎ from 
(
  select m.S [学子编号] , 
         m.Sname [学员姓名] ,
         isnull(cast(avg(score) as decimal(18,2)),0) [平均战绩]
  from Student m left join SC n on m.S = n.S 
  group by m.S , m.Sname
State of Qatar t2 where 平均战绩 >= t1.等分成绩卡塔尔国 from 
(
  select m.S [学员编号] , 
         m.Sname [学子姓名] ,
         isnull(cast(avg(score) as decimal(18,2)),0) [平均战表]
  from Student m left join SC n on m.S = n.S 
  group by m.S , m.Sname
) t1
order by px
–24.2 查询学子的平均成绩并开展排行,sql
二〇〇七用rank,DENSE_RANK完毕,分平均战表再一次时保留排行空缺和不保留排行空缺二种。
select t.* , px = rank() over(order by [平均成绩] desc) from
(
  select m.S [学员编号] , 
         m.Sname [学子姓名] ,
         isnull(cast(avg(score) as decimal(18,2)),0) [平均战表]
  from Student m left join SC n on m.S = n.S 
  group by m.S , m.Sname
) t
order by px

select t.* , px = DENSE_RANK() over(order by [平均成绩] desc) from
(
  select m.S [学子编号] , 
         m.Sname [学员姓名] ,
         isnull(cast(avg(score) as decimal(18,2)),0) [平均成绩]
  from Student m left join SC n on m.S = n.S 
  group by m.S , m.Sname
) t
order by px
  
–25、查询各科成绩前三名的记录
–25.1 分数重复时保留排行空缺
select m.* , n.C , n.score from Student m, SC n where m.S = n.S and
n.score in 
(select top 3 score from sc where C = n.C order by score desc) order by
n.C , n.score desc
–25.2 分数重复时不保留名次空缺,合併排行
–sql 2003用子查询完毕
select * from (select t.* , px = (select count(distinct score) from SC
where C = t.C and score >= t.score) from sc t) m where px between 1
and 3 order by m.C , m.px 
–sql 2005用DENSE_RANK实现
select * from (select t.* , px = DENSE_RANK() over(partition by C
order by score desc) from sc t) m where px between 1 and 3 order by m.C
, m.px

–26、查询每门学科被选修的学习者数 
select C , count(S)[学生数] from sc group by C

–27、查询出唯有两门学科的全套学员的学号和姓名 
select Student.S , Student.Sname
from Student , SC 
where Student.S = SC.S 
group by Student.S , Student.Sname
having count(SC.C) = 2
order by Student.S
 
–28、查询男人、女孩子人数 
select count(Ssex卡塔尔(قطر‎ as 男人人数 from Student where Ssex = N’男’
select count(Ssex卡塔尔(قطر‎ as 女人人数 from Student where Ssex = N’女’
select sum(case when Ssex = N’男’ then 1 else 0 end)
[男士人数],sum(case when Ssex = N’女’ then 1 else 0 end) [女子人数]
from student
select case when Ssex = N’男’ then N’男生人数’ else N’女孩子人数’ end
[子女情事] , count(1) [人数] from student group by case when Ssex =
N’男’ then N’男人人数’ else N’女人人数’ end

–29、查询名字中蕴藏”风”字的学习者新闻
select * from student where sname like N’%风%’
select * from student where charindex(N’风’ , sname) > 0

–30、查询同名同种性别学子名单,并总计同有名的人数 
select Sname [学子姓名], count(*) [人数] from Student group by
Sname having count(*) > 1
 
–31、查询壹玖捌玖年落榜的学子名单(注:Student表中Sage列的连串是datetime卡塔尔(قطر‎ 
select * from Student where year(sage) = 1990
select * from Student where datediff(yy,sage,’1990-01-01′) = 0
select * from Student where datepart(yy,sage) = 1990
select * from Student where convert(varchar(4),sage,120) = ‘1990’

–32、查询每门课程的平分成绩,结果按平均战绩降序排列,平均战表相像时,按学科编号升序排列 
select m.C , m.Cname , cast(avg(n.score) as decimal(18,2)) avg_score
from Course m, SC n 
where m.C = n.C    
group by m.C , m.Cname 
order by avg_score desc, m.C asc

–33、查询平均成绩超越等于85的具备学员的学号、姓名和平均成绩 
select a.S , a.Sname , cast(avg(b.score) as decimal(18,2)) avg_score
from Student a , sc b
where a.S = b.S
group by a.S , a.Sname
having cast(avg(b.score) as decimal(18,2)) >= 85 
order by a.S

–34、查询课程名称叫”数学”,且分数低于60的学生姓名和分数 
select sname , score
from Student , SC , Course 
where SC.S = Student.S and SC.C = Course.C and Course.Cname = N’数学’
and score < 60

–35、查询全数学员的教程及分数情状; 
select Student.* , Course.Cname , SC.C , SC.score  
from Student, SC , Course 
where Student.S = SC.S and SC.C = Course.C 
order by Student.S , SC.C

–36、查询任何一门学科战表在70分以上的姓名、课程名称和分数; 
select Student.* , Course.Cname , SC.C , SC.score  
from Student, SC , Course 
where Student.S = SC.S and SC.C = Course.C and SC.score >= 70 
order by Student.S , SC.C

–37、查询比不上格的教程
select Student.* , Course.Cname , SC.C , SC.score  
from Student, SC , Course 
where Student.S = SC.S and SC.C = Course.C and SC.score < 60 
order by Student.S , SC.C

–38、查询课程编号为01且课程战绩在80分以上的学习者的学号和人名; 
select Student.* , Course.Cname , SC.C , SC.score  
from Student, SC , Course 
where Student.S = SC.S and SC.C = Course.C and SC.C = ’01’ and SC.score
>= 80 
order by Student.S , SC.C

–39、求每门学科的学习者人数 
select Course.C , Course.Cname , count(*) [学员人数]
from Course , SC 
where Course.C = SC.C
group by  Course.C , Course.Cname
order by Course.C , Course.Cname

–40、查询选修”张三”老师所授课程的学员中,成绩最高的学员消息及其成绩
–40.1 当最高分独有三个时
select top 1 Student.* , Course.Cname , SC.C , SC.score  
from Student, SC , Course , Teacher
where Student.S = SC.S and SC.C = Course.C and Course.T = Teacher.T and
Teacher.Tname = N’张三’
order by SC.score desc
–40.2 当最高分现身八个时
select Student.* , Course.Cname , SC.C , SC.score  
from Student, SC , Course , Teacher
where Student.S = SC.S and SC.C = Course.C and Course.T = Teacher.T and
Teacher.Tname = N’张三’ and
SC.score = (select max(SC.score) from SC , Course , Teacher where SC.C =
Course.C and Course.T = Teacher.T and Teacher.Tname = N’张三’)

–41、查询不一致科目成绩形似的学习者的学习者编号、课程编号、学生成绩 
–方法1
select m.* from SC m ,(select C , score from SC group by C , score
having count(1) > 1) n 
where m.C= n.C and m.score = n.score order by m.C , m.score , m.S
–方法2
select m.* from SC m where exists (select 1 from (select C , score from
SC group by C , score having count(1) > 1) n 
where m.C= n.C and m.score = n.score) order by m.C , m.score , m.S

–42、查询每门功成绩最棒的前两名 
select t.* from sc t where score in (select top 2 score from sc where C
= T.C order by score desc) order by t.C , t.score desc

–43、总结每门课程的学子选修人数(超越5人的科目才总括)。必要输出课程号和选修人数,查询结果按人头降序排列,若人数相似,按学科号升序排列  
select Course.C , Course.Cname , count(*) [学子人数]
from Course , SC 
where Course.C = SC.C
group by  Course.C , Course.Cname
having count(*) >= 5
order by [学子人数] desc , Course.C

–44、检索最少选修两门学科的学习者学号 
select student.S , student.Sname 
from student , SC 
where student.S = SC.S 
group by student.S , student.Sname 
having count(1) >= 2
order by student.S

–45、查询选修了全副学科的上学的小孩子音讯 
–方法1 依据数据来实现
select student.* from student where S in
(select S from sc group by S having count(1) = (select count(1) from
course))
–方法2 用到重复否定来完结
select t.* from student t where t.S not in 
(
  select distinct m.S from
  (
    select S , C from student , course 
  ) m where not exists (select 1 from sc n where n.S = m.S and n.C =
m.C)
)
–方法3 利用重复否定来产生
select t.* from student t where not exists(select 1 from 
(
  select distinct m.S from
  (
    select S , C from student , course 
  ) m where not exists (select 1 from sc n where n.S = m.S and n.C =
m.C)
) k where k.S = t.S
)

–46、查询各学子的年纪
–46.1 只根据年度来算
select * , datediff(yy , sage , getdate()) [年龄] from student
–46.2 依照出生日期来算,当前月日 < 出生年月的月日则,年龄减一
select * , case when right(convert(varchar(10),getdate(),120),5) <
right(convert(varchar(10),sage,120),5) then datediff(yy , sage ,
getdate()) – 1 else datediff(yy , sage , getdate()) end [年龄] from
student

–47、查询本周过华诞的上学的儿童
select * from student where datediff(week,datename(yy,getdate()) +
right(convert(varchar(10),sage,120),6),getdate()) = 0

–48、查询上周过华诞的学员
select * from student where datediff(week,datename(yy,getdate()) +
right(convert(varchar(10),sage,120),6),getdate()) = -1

–49、查询上个月过生辰的学子
select * from student where datediff(mm,datename(yy,getdate()) +
right(convert(varchar(10),sage,120),6),getdate()) = 0

–50、查询下一个月过寿辰的学习者
select * from student where datediff(mm,datename(yy,getdate()) +
right(convert(varchar(10),sage,120),6),getdate()) = -1

drop table  Student,Course,Teacher,SC

–1.学生表
Student(S,Sname,Sage,Ssex卡塔尔 –S 学子编号,Sname 学子姓名,Sage
出生年月,Ssex 学子性别
–2.课程表 
Course(C,Cname,T卡塔尔 –C –课程编号,Cname 课程名称,T 教师编号
–3.教师表 
Teacher(T,Tname卡塔尔国 –T 教师编号,Tname 教授姓名
–4.成绩表 
SC(S,C,score卡塔尔国 –S 学子编号,C 课程编号,score 分数
*/
–创制测量试验数据
create table Student(S varchar(10),Sname nvarchar(10),Sage datetime,Ssex
nvarchar(10))
insert into Student values(’01’ , N’赵雷’ , ‘1990-01-01′ , N’男’)
insert into Student values(’02’ , N’钱电’ , ‘1990-12-21′ , N’男’)
insert into Student values(’03’ , N’孙风’ , ‘1990-05-20′ , N’男’)
insert into Student values(’04’ , N’李云’ , ‘1990-08-06′ , N’男’)
insert into Student values(’05’ , N’周梅’ , ‘1991-12-01′ , N’女’)
insert into Student values(’06’ , N’吴兰’ , ‘1992-03-01′ , N’女’)
insert into Student values(’07’ , N’郑竹’ , ‘1989-07-01′ , N’女’)
insert into Student values(’08’ , N’王菊’ , ‘1990-01-20′ , N’女’)
create table Course(C varchar(10),Cname nvarchar(10),T varchar(10))
insert into Course values(’01’ , N’语文’ , ’02’)
insert into Course values(’02’ , N’数学’ , ’01’)
insert into Course values(’03’ , N’英语’ , ’03’)
create table Teacher(T varchar(10),Tname nvarchar(10))
insert into Teacher values(’01’ , N’张三’)
insert into Teacher values(’02’ , N’李四’)
insert into Teacher values(’03’ , N’王五’)
create table SC(S varchar(10),C varchar(10),score decimal(18,1))
insert into SC values(’01’ , ’01’ , 80)
insert into SC values(’01’ , ’02’ , 90)
insert into SC values(’01’ , ’03’ , 99)
insert into SC values(’02’ , ’01’ , 70)
insert into SC values(’02’ , ’02’ , 60)
insert into SC values(’02’ , ’03’ , 80)
insert into SC values(’03’ , ’01’ , 80)
insert into SC values(’03’ , ’02’ , 80)
insert into SC values(’03’ , ’03’ , 80)
insert into SC values(’04’ , ’01’ , 50)
insert into SC values(’04’ , ’02’ , 30)
insert into SC values(’04’ , ’03’ , 20)
insert into SC values(’05’ , ’01’ , 76)
insert into SC values(’05’ , ’02’ , 87)
insert into SC values(’06’ , ’01’ , 31)
insert into SC values(’06’ , ’03’ , 34)
insert into SC values(’07’ , ’02’ , 89)
insert into SC values(’07’ , ’03’ , 98)
go

–1、查询”01″课程比”02″课程成绩高的学子的消息及学科分数
–1.1、查询同期设有”01″课程和”02″课程的景观
select a.* , b.score [课程’01’的分数],c.score [课程’02’的分数] from
Student a , SC b , SC c 
where a.S = b.S and a.S = c.S and b.C = ’01’ and c.C = ’02’ and b.score
> c.score
–1.2、查询相同的时间存在”01″课程和”02″课程的情事和存在”01″课程但也许空中楼阁”02″课程的情况(不设有的时候显得为null卡塔尔国(以下存在相近内容时不再解释卡塔尔(قطر‎
select a.* , b.score [课程”01″的分数],c.score [课程”02″的分数] from
Student a 
left join SC b on a.S = b.S and b.C = ’01’
left join SC c on a.S = c.S and c.C = ’02’
where b.score > isnull(c.score,0)

–2、查询”01″课程比”02″课程战表低的学员的音信及学科分数
–2.1、查询同不常候设有”01″课程和”02″课程的意况
select a.* , b.score [课程’01’的分数],c.score [课程’02’的分数] from
Student a , SC b , SC c 
where a.S = b.S and a.S = c.S and b.C = ’01’ and c.C = ’02’ and b.score
< c.score
–2.2、查询同一时候设有”01″课程和”02″课程的状态和海市蜃楼”01″课程但存在”02″课程的景况
select a.* , b.score [课程”01″的分数],c.score [课程”02″的分数] from
Student a 
left join SC b on a.S = b.S and b.C = ’01’
left join SC c on a.S = c.S and c.C = ’02’
where isnull(b.score,0) < c.score

–3、查询平均成绩超越等于60分的同桌的学习者编号和学员姓名和平均战表
select a.S , a.Sname , cast(avg(b.score) as decimal(18,2)) avg_score
from Student a , sc b
where a.S = b.S
group by a.S , a.Sname
having cast(avg(b.score) as decimal(18,2)) >= 60 
order by a.S

–4、查询平均战表小于60分的同班的学员编号和学子姓名和平均成绩
–4.1、查询在sc表存在战表的学员音讯的SQL语句。
select a.S , a.Sname , cast(avg(b.score) as decimal(18,2)) avg_score
from Student a , sc b
where a.S = b.S
group by a.S , a.Sname
having cast(avg(b.score) as decimal(18,2)) < 60 
order by a.S
–4.2、查询在sc表中不设有成绩的学员音信的SQL语句。
select a.S , a.Sname , isnull(cast(avg(b.score) as decimal(18,2)),0)
avg_score
from Student a left join sc b
on a.S = b.S
group by a.S , a.Sname
having isnull(cast(avg(b.score) as decimal(18,2)),0) < 60 
order by a.S

–5、查询全数同学的学员编号、学子姓名、选课总量、全数科指标总成绩
–5.1、查询全部有战表的SQL。
select a.S [学员编号], a.Sname [学员姓名], count(b.C卡塔尔(قطر‎ 选课总的数量,
sum(score卡塔尔(قطر‎ [持有科目的总战绩]
from Student a , SC b 
where a.S = b.S 
group by a.S,a.Sname 
order by a.S
–5.2、查询全部(包含有战绩和无成绩卡塔尔的SQL。
select a.S [学员编号], a.Sname [学员姓名], count(b.CState of Qatar 选课总量,
sum(score卡塔尔 [具有科目标总成绩]
from Student a left join SC b 
on a.S = b.S 
group by a.S,a.Sname 
order by a.S

–6、查询”李”姓老师的数目 
–方法1
select count(Tname) [“李”姓老师的数额] from Teacher where Tname like
N’李%’
–方法2
select count(Tname) [“李”姓老师的数据] from Teacher where
left(Tname,1) = N’李’
/*
“李”姓老师的数目   
———– 
1
*/

–7、查询学过”张三”老师上课的同桌的音信 
select distinct Student.* from Student , SC , Course , Teacher 
where Student.S = SC.S and SC.C = Course.C and Course.T = Teacher.T and
Teacher.Tname = N’张三’
order by Student.S

–8、查询没学过”张三”老师解说的同学的新闻 
select m.* from Student m where S not in (select distinct SC.S from SC
, Course , Teacher where SC.C = Course.C and Course.T = Teacher.T and
Teacher.Tname = N’张三’) order by m.S

–9、查询学过数码为”01″况且也学过数码为”02″的教程的同校的音讯
–方法1
select Student.* from Student , SC where Student.S = SC.S and SC.C =
’01’ and exists (Select 1 from SC SC_2 where SC_2.S = SC.S and SC_2.C
= ’02’) order by Student.S
–方法2
select Student.* from Student , SC where Student.S = SC.S and SC.C =
’02’ and exists (Select 1 from SC SC_2 where SC_2.S = SC.S and SC_2.C
= ’01’) order by Student.S
–方法3
select m.* from Student m where S in
(
  select S from
  (
    select distinct S from SC where C = ’01’
    union all
    select distinct S from SC where C = ’02’
  ) t group by S having count(1) = 2 
)
order by m.S

–10、查询学过数码为”01″然则并未学过数码为”02″的学科的校友的新闻
–方法1
select Student.* from Student , SC where Student.S = SC.S and SC.C =
’01’ and not exists (Select 1 from SC SC_2 where SC_2.S = SC.S and
SC_2.C = ’02’) order by Student.S
–方法2
select Student.* from Student , SC where Student.S = SC.S and SC.C =
’01’ and Student.S not in (Select SC_2.S from SC SC_2 where SC_2.S =
SC.S and SC_2.C = ’02’) order by Student.S

–11、查询未有学全全部课程的同室的新闻 
–11.1、
select Student.*
from Student , SC 
where Student.S = SC.S 
group by Student.S , Student.Sname , Student.Sage , Student.Ssex having
count(C) < (select count(C) from Course) 
–11.2
select Student.*
from Student left join SC 
on Student.S = SC.S 
group by Student.S , Student.Sname , Student.Sage , Student.Ssex having
count(C) < (select count(C) from Course)

–12、查询至稀有一门课与学号为”01″的校友所学相符的同桌的音信 
select distinct Student.* from Student , SC where Student.S = SC.S and
SC.C in (select C from SC where S = ’01’) and Student.S <> ’01’

–13、查询和”01″号的同学学习的教程完全相像的别的同学的信息 
select Student.* from Student where S in
(select distinct SC.S from SC where S <> ’01’ and SC.C in (select
distinct C from SC where S = ’01’) 
group by SC.S having count(1) = (select count(1) from SC where S=’01’))

–14、查询没学过”张三”老师教师的任一门学科的学子姓名 
select student.* from student where student.S not in 
(select distinct sc.S from sc , course , teacher where sc.C = course.C
and course.T = teacher.T and teacher.tname = N’张三’)
order by student.S

–15、查询两门及其以上不如格课程的同窗的学号,姓名及其平均成绩 
select student.S , student.sname , cast(avg(score) as decimal(18,2))
avg_score from student , sc 
where student.S = SC.S and student.S in (select S from SC where score
< 60 group by S having count(1) >= 2)
group by student.S , student.sname

–16、检索”01″课程分数小于60,按分数降序排列的学子消息
select student.* , sc.C , sc.score from student , sc 
where student.S = SC.S and sc.score < 60 and sc.C = ’01’
order by sc.score desc 

–17、按平均战绩从高到低展现全体学子的富有课程的大成以致平均成绩
–17.1 SQL 2000 静态 
select a.S 学子编号 , a.Sname 学生姓名 ,
       max(case c.Cname when N’语文’ then b.score else null end)
[语文],
       max(case c.Cname when N’数学’ then b.score else null end)
[数学],
       max(case c.Cname when N’英语’ then b.score else null end)
[英语],
       cast(avg(b.score) as decimal(18,2)) 平均分
from Student a 
left join SC b on a.S = b.S
left join Course c on b.C = c.C
group by a.S , a.Sname
order by 平均分 desc
–17.2 SQL 2000 动态 
declare @sql nvarchar(4000)
set @sql = ‘select a.S ‘ + N’学子编号’ + ‘ , a.Sname ‘ + N’学子姓名’
select @sql = @sql + ‘,max(case c.Cname when N”’+Cname+”’ then b.score
else null end) [‘+Cname+’]’
from (select distinct Cname from Course) as t
set @sql = @sql + ‘ , cast(avg(b.score) as decimal(18,2)) ‘ + N’平均分’

  • ‘ from Student a left join SC b on a.S = b.S left join Course c on b.C
    = c.C
    group by a.S , a.Sname order by ‘ + N’平均分’ + ‘ desc’
    exec(@sql)
    –17.3 有关sql 二〇〇六的动静态写法参见小编的篇章《普通行列调换(version
    2.0卡塔尔(قطر‎》或《普通行列调换(version 3.0State of Qatar》。

 

对自己有用[9]丢个板砖[0]援引举报管理TOP 回复次数:1043

dawugui
(爱新觉罗.毓华State of Qatar
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更加多勋章
#1楼 得分:0回复于:2010-05-17 17:47:07
SQL code
–18、查询各科战表最高分、最低分和平均分:以如下方式呈现:课程ID,课程name,最高分,最低分,平均分,及格率,中等率,非凡率,杰出率
–及格为>=60,中等为:70-80,优良为:80-90,优秀为:>=90
–方法1
select m.C [课程编号], m.Cname [课程名称], 
  max(n.score) [最高分],
  min(n.score) [最低分],
  cast(avg(n.score) as decimal(18,2)) [平均分],
  cast((select count(1) from SC where C = m.C and score >= 60)*100.0
/ (select count(1) from SC where C = m.C) as decimal(18,2))
[及格率(%)],
  cast((select count(1) from SC where C = m.C and score >= 70 and
score < 80 )*100.0 / (select count(1) from SC where C = m.C) as
decimal(18,2)) [中等率(%)],
  cast((select count(1) from SC where C = m.C and score >= 80 and
score < 90 )*100.0 / (select count(1) from SC where C = m.C) as
decimal(18,2)) [优良率(%)],
  cast((select count(1) from SC where C = m.C and score >= 90)*100.0
/ (select count(1) from SC where C = m.C) as decimal(18,2))
[优秀率(%)]
from Course m , SC n
where m.C = n.C
group by m.C , m.Cname
order by m.C
–方法2
select m.C [学科编号], m.Cname [课程名称], 
  (select max(score) from SC where C = m.C) [最高分],
  (select min(score) from SC where C = m.C) [最低分],
  (select cast(avg(score) as decimal(18,2)) from SC where C = m.C)
[平均分],
  cast((select count(1) from SC where C = m.C and score >= 60)*100.0
/ (select count(1) from SC where C = m.C) as decimal(18,2))
[及格率(%)],
  cast((select count(1) from SC where C = m.C and score >= 70 and
score < 80 )*100.0 / (select count(1) from SC where C = m.C) as
decimal(18,2)) [中等率(%)],
  cast((select count(1) from SC where C = m.C and score >= 80 and
score < 90 )*100.0 / (select count(1) from SC where C = m.C) as
decimal(18,2)) [优良率(%)],
  cast((select count(1) from SC where C = m.C and score >= 90)*100.0
/ (select count(1) from SC where C = m.C) as decimal(18,2))
[优秀率(%)]
from Course m 
order by m.C

–19、按各科成绩进行排序,并出示排行
–19.1 sql 二零零一用子查询实现
–Score重复时保留排名空缺
select t.* , px = (select count(1) from SC where C = t.C and score >
t.score) + 1 from sc t order by t.C , px 
–Score重复时合併排行
select t.* , px = (select count(distinct score) from SC where C = t.C
and score >= t.score) from sc t order by t.C , px 
–19.2 sql 2005用rank,DENSE_RANK完成
–Score重复时保留排名空缺(rank完结卡塔尔(قطر‎
select t.* , px = rank() over(partition by C order by score desc) from
sc t order by t.C , px 
–Score重复时合併排名(DENSE_RANK完成)
select t.* , px = DENSE_RANK() over(partition by C order by score
desc) from sc t order by t.C , px

–20、查询学子的总战绩并打开排名
–20.1 查询学子的总成绩
select m.S [学子编号] , 
       m.Sname [学子姓名] ,
       isnull(sum(score),0) [总成绩]
from Student m left join SC n on m.S = n.S 
group by m.S , m.Sname
order by [总成绩] desc
–20.2 查询学子的总成绩并拓宽排行,sql
二〇〇二用子查询完结,分总分重复时保留排名空缺和不保留排名空缺二种。
select t1.* , px = (select count(1) from 
(
  select m.S [学子编号] , 
         m.Sname [学生姓名] ,
         isnull(sum(score),0) [总成绩]
  from Student m left join SC n on m.S = n.S 
  group by m.S , m.Sname
) t2 where 总成绩 > t1.总成绩) + 1 from 
(
  select m.S [学子编号] , 
         m.Sname [学员姓名] ,
         isnull(sum(score),0) [总成绩]
  from Student m left join SC n on m.S = n.S 
  group by m.S , m.Sname
) t1
order by px

select t1.* , px = (select count(distinct 总成绩) from 
(
  select m.S [学员编号] , 
         m.Sname [学员姓名] ,
         isnull(sum(score),0) [总成绩]
  from Student m left join SC n on m.S = n.S 
  group by m.S , m.Sname
) t2 where 总成绩 >= t1.总成绩) from 
(
  select m.S [学员编号] , 
         m.Sname [学子姓名] ,
         isnull(sum(score),0) [总成绩]
  from Student m left join SC n on m.S = n.S 
  group by m.S , m.Sname
) t1
order by px
–20.3 查询学子的总成绩并拓宽排行,sql
二零零七用rank,DENSE_RANK达成,分总分重复时保留排名空缺和不保留排行空缺两种。
select t.* , px = rank() over(order by [总成绩] desc) from
(
  select m.S [学子编号] , 
         m.Sname [学子姓名] ,
         isnull(sum(score),0) [总成绩]
  from Student m left join SC n on m.S = n.S 
  group by m.S , m.Sname
) t
order by px

select t.* , px = DENSE_RANK() over(order by [总成绩] desc) from
(
  select m.S [学子编号] , 
         m.Sname [学子姓名] ,
         isnull(sum(score),0) [总成绩]
  from Student m left join SC n on m.S = n.S 
  group by m.S , m.Sname
) t
order by px

–21、查询差别老师所教差别科目平均分从高到低突显 
select m.T , m.Tname , cast(avg(o.score) as decimal(18,2)) avg_score
from Teacher m , Course n , SC o
where m.T = n.T and n.C = o.C
group by m.T , m.Tname
order by avg_score desc

–22、查询全数课程的实际业绩第2名到第3名的学习者音讯及该科目战表
–22.1 sql 2003用子查询完结
–Score重复时保留排行空缺
select * from (select t.* , px = (select count(1) from SC where C =
t.C and score > t.score) + 1 from sc t) m where px between 2 and 3
order by m.C , m.px 
–Score重复时合併排行
select * from (select t.* , px = (select count(distinct score) from SC
where C = t.C and score >= t.score) from sc t) m where px between 2
and 3 order by m.C , m.px 
–22.2 sql 2005用rank,DENSE_RANK完成
–Score重复时保留排名空缺(rank达成卡塔尔
select * from (select t.* , px = rank() over(partition by C order by
score desc) from sc t) m where px between 2 and 3 order by m.C , m.px 
–Score重复时合併排名(DENSE_RANK完成)
select * from (select t.* , px = DENSE_RANK() over(partition by C
order by score desc) from sc t) m where px between 2 and 3 order by m.C
, m.px

–23、总结各科战绩各分数段人数:课程编号,课程名称,[100-85],[85-70],[70-60],[0-60]及所占百分比 
–23.1
总计各科战表各分数段人数:课程编号,课程名称,[100-85],[85-70],[70-60],[0-60]
–横向彰显
select Course.C [学科编号] , Cname as [课程名称] ,
  sum(case when score >= 85 then 1 else 0 end) [85-100],
  sum(case when score >= 70 and score < 85 then 1 else 0 end)
[70-85],
  sum(case when score >= 60 and score < 70 then 1 else 0 end)
[60-70],
  sum(case when score < 60 then 1 else 0 end) [0-60]
from sc , Course 
where SC.C = Course.C 
group by Course.C , Course.Cname
order by Course.C
–纵向呈现1(展现存在的分数段State of Qatar
select m.C [学科编号] , m.Cname [课程名称] , 分数段 = (
  case when n.score >= 85 then ’85-100′
       when n.score >= 70 and n.score < 85 then ’70-85′
       when n.score >= 60 and n.score < 70 then ’60-70′
       else ‘0-60’
  end) , 
  count(1) 数量 
from Course m , sc n
where m.C = n.C 
group by m.C , m.Cname , (
  case when n.score >= 85 then ’85-100′
       when n.score >= 70 and n.score < 85 then ’70-85′
       when n.score >= 60 and n.score < 70 then ’60-70′
       else ‘0-60’
  end)
order by m.C , m.Cname , 分数段
–纵向展现2(展现存在的分数段,不设有的分数段用0突显State of Qatar
select m.C [课程编号] , m.Cname [课程名称] , 分数段 = (
  case when n.score >= 85 then ’85-100′
       when n.score >= 70 and n.score < 85 then ’70-85′
       when n.score >= 60 and n.score < 70 then ’60-70′
       else ‘0-60’
  end) , 
  count(1) 数量 
from Course m , sc n
where m.C = n.C 
group by all m.C , m.Cname , (
  case when n.score >= 85 then ’85-100′
       when n.score >= 70 and n.score < 85 then ’70-85′
       when n.score >= 60 and n.score < 70 then ’60-70′
       else ‘0-60’
  end)
order by m.C , m.Cname , 分数段

–23.2
总计各科成绩各分数段人数:课程编号,课程名称,[100-85],[85-70],[70-60],[<60]及所占百分比 
–横向展现
select m.C 课程编号, m.Cname 课程名称,
  (select count(1) from SC where C = m.C and score < 60) [0-60],
  cast((select count(1) from SC where C = m.C and score < 60)*100.0
/ (select count(1) from SC where C = m.C) as decimal(18,2))
[百分比(%)],
  (select count(1) from SC where C = m.C and score >= 60 and score
< 70) [60-70],
  cast((select count(1) from SC where C = m.C and score >= 60 and
score < 70)*100.0 / (select count(1) from SC where C = m.C) as
decimal(18,2)) [百分比(%)],
  (select count(1) from SC where C = m.C and score >= 70 and score
< 85) [70-85],
  cast((select count(1) from SC where C = m.C and score >= 70 and
score < 85)*100.0 / (select count(1) from SC where C = m.C) as
decimal(18,2)) [百分比(%)],
  (select count(1) from SC where C = m.C and score >= 85)
[85-100],
  cast((select count(1) from SC where C = m.C and score >= 85)*100.0
/ (select count(1) from SC where C = m.C) as decimal(18,2))
[百分比(%)]
from Course m 
order by m.C
–纵向展现1(展现存在的分数段卡塔尔(قطر‎
select m.C [学科编号] , m.Cname [课程名称] , 分数段 = (
  case when n.score >= 85 then ’85-100′
       when n.score >= 70 and n.score < 85 then ’70-85′
       when n.score >= 60 and n.score < 70 then ’60-70′
       else ‘0-60’
  end) , 
  count(1) 数量 ,  
  cast(count(1) * 100.0 / (select count(1) from sc where C = m.C) as
decimal(18,2)) [百分比(%)]
from Course m , sc n
where m.C = n.C 
group by m.C , m.Cname , (
  case when n.score >= 85 then ’85-100′
       when n.score >= 70 and n.score < 85 then ’70-85′
       when n.score >= 60 and n.score < 70 then ’60-70′
       else ‘0-60’
  end)
order by m.C , m.Cname , 分数段
–纵向突显2(显示存在的分数段,不设有的分数段用0显示卡塔尔国
select m.C [课程编号] , m.Cname [课程名称] , 分数段 = (
  case when n.score >= 85 then ’85-100′
       when n.score >= 70 and n.score < 85 then ’70-85′
       when n.score >= 60 and n.score < 70 then ’60-70′
       else ‘0-60’
  end) , 
  count(1) 数量 ,  
  cast(count(1) * 100.0 / (select count(1) from sc where C = m.C) as
decimal(18,2)) [百分比(%)]
from Course m , sc n
where m.C = n.C 
group by all m.C , m.Cname , (
  case when n.score >= 85 then ’85-100′
       when n.score >= 70 and n.score < 85 then ’70-85′
       when n.score >= 60 and n.score < 70 then ’60-70′
       else ‘0-60’
  end)
order by m.C , m.Cname , 分数段

 

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精髓推荐:SQL语句优化汇总

dawugui
(爱新觉罗.毓华卡塔尔国
等 级:
3
越多勋章
#2楼 得分:0回复于:2010-05-17 17:47:22
SQL code
–24、查询学毕生均战绩及其排名 
–24.1 查询学生的平分战表并张开排名,sql
二零零二用子查询完成,分平均成绩再一次时保留排行空缺和不保留排名空缺三种。
select t1.* , px = (select count(1) from 
(
  select m.S [学员编号] , 
         m.Sname [学员姓名] ,
         isnull(cast(avg(score) as decimal(18,2)),0) [平均战表]
  from Student m left join SC n on m.S = n.S 
  group by m.S , m.Sname
卡塔尔国 t2 where 平均成绩 > t1.等分成绩State of Qatar + 1 from 
(
  select m.S [学子编号] , 
         m.Sname [学子姓名] ,
         isnull(cast(avg(score) as decimal(18,2)),0) [平均成绩]
  from Student m left join SC n on m.S = n.S 
  group by m.S , m.Sname
) t1
order by px

select t1.* , px = (select count(distinct 平均战表卡塔尔国 from 
(
  select m.S [学员编号] , 
         m.Sname [学子姓名] ,
         isnull(cast(avg(score) as decimal(18,2)),0) [平均战表]
  from Student m left join SC n on m.S = n.S 
  group by m.S , m.Sname
卡塔尔国 t2 where 平均成绩 >= t1.平均战绩State of Qatar from 
(
  select m.S [学员编号] , 
         m.Sname [学员姓名] ,
         isnull(cast(avg(score) as decimal(18,2)),0) [平均战表]
  from Student m left join SC n on m.S = n.S 
  group by m.S , m.Sname
) t1
order by px
–24.2 查询学子的平分成绩并进行排名,sql
二零零五用rank,DENSE_RANK实现,分平均战表再一次时保留排行空缺和不保留排名空缺三种。
select t.* , px = rank() over(order by [平均战绩] desc) from
(
  select m.S [学子编号] , 
         m.Sname [学员姓名] ,
         isnull(cast(avg(score) as decimal(18,2)),0) [平均战表]
  from Student m left join SC n on m.S = n.S 
  group by m.S , m.Sname
) t
order by px

select t.* , px = DENSE_RANK() over(order by [平均战表] desc) from
(
  select m.S [学员编号] , 
         m.Sname [学生姓名] ,
         isnull(cast(avg(score) as decimal(18,2)),0) [平均成绩]
  from Student m left join SC n on m.S = n.S 
  group by m.S , m.Sname
) t
order by px
  
–25、查询各科战绩前三名的笔录
–25.1 分数重复时保留排名空缺
select m.* , n.C , n.score from Student m, SC n where m.S = n.S and
n.score in 
(select top 3 score from sc where C = n.C order by score desc) order by
n.C , n.score desc
–25.2 分数重复时不保留排行空缺,归总排行
–sql 2002用子查询实现
select * from (select t.* , px = (select count(distinct score) from SC
where C = t.C and score >= t.score) from sc t) m where px between 1
and 3 order by m.C , m.px 
–sql 2005用DENSE_RANK实现
select * from (select t.* , px = DENSE_RANK() over(partition by C
order by score desc) from sc t) m where px between 1 and 3 order by m.C
, m.px

–26、查询每门课程被选修的上学的小孩子数 
select C , count(S)[学生数] from sc group by C

–27、查询出唯有两门课程的上上下下学子的学号和姓名 
select Student.S , Student.Sname
from Student , SC 
where Student.S = SC.S 
group by Student.S , Student.Sname
having count(SC.C) = 2
order by Student.S
 
–28、查询男人、女人人数 
select count(SsexState of Qatar as 男人人数 from Student where Ssex = N’男’
select count(SsexState of Qatar as 女孩子人数 from Student where Ssex = N’女’
select sum(case when Ssex = N’男’ then 1 else 0 end)
[男士人数],sum(case when Ssex = N’女’ then 1 else 0 end) [女子人数]
from student
select case when Ssex = N’男’ then N’男人人数’ else N’女子人数’ end
[男女情事] , count(1) [人数] from student group by case when Ssex =
N’男’ then N’男生人数’ else N’女人人数’ end

–29、查询名字中带有”风”字的上学的小孩子音讯
select * from student where sname like N’%风%’
select * from student where charindex(N’风’ , sname) > 0

–30、查询同名同性别学子名单,并总括同有名气的人数 
select Sname [学员姓名], count(*) [人数] from Student group by
Sname having count(*) > 1
 
–31、查询1987年降生的学子名单(注:Student表中Sage列的类型是datetime卡塔尔国 
select * from Student where year(sage) = 1990
select * from Student where datediff(yy,sage,’1990-01-01′) = 0
select * from Student where datepart(yy,sage) = 1990
select * from Student where convert(varchar(4),sage,120) = ‘1990’

–32、查询每门科目标平均战表,结果按平均战表降序排列,平均战表同样时,按学科编号升序排列 
select m.C , m.Cname , cast(avg(n.score) as decimal(18,2)) avg_score
from Course m, SC n 
where m.C = n.C    
group by m.C , m.Cname 
order by avg_score desc, m.C asc

–33、查询平均战绩当先等于85的全数学子的学号、姓名和平均战表 
select a.S , a.Sname , cast(avg(b.score) as decimal(18,2)) avg_score
from Student a , sc b
where a.S = b.S
group by a.S , a.Sname
having cast(avg(b.score) as decimal(18,2)) >= 85 
order by a.S

–34、查询课程名为”数学”,且分数低于60的学习者姓名和分数 
select sname , score
from Student , SC , Course 
where SC.S = Student.S and SC.C = Course.C and Course.Cname = N’数学’
and score < 60

–35、查询全体学生的课程及分数情状; 
select Student.* , Course.Cname , SC.C , SC.score  
from Student, SC , Course 
where Student.S = SC.S and SC.C = Course.C 
order by Student.S , SC.C

–36、查询任何一门课程成绩在70分以上的人名、课程名称和分数; 
select Student.* , Course.Cname , SC.C , SC.score  
from Student, SC , Course 
where Student.S = SC.S and SC.C = Course.C and SC.score >= 70 
order by Student.S , SC.C

–37、查询不比格的课程
select Student.* , Course.Cname , SC.C , SC.score  
from Student, SC , Course 
where Student.S = SC.S and SC.C = Course.C and SC.score < 60 
order by Student.S , SC.C

–38、查询课程编号为01且课程战绩在80分以上的上学的小孩子的学号和姓名; 
select Student.* , Course.Cname , SC.C , SC.score  
from Student, SC , Course 
where Student.S = SC.S and SC.C = Course.C and SC.C = ’01’ and SC.score
>= 80 
order by Student.S , SC.C

–39、求每门科目标学员人数 
select Course.C , Course.Cname , count(*) [学员人数]
from Course , SC 
where Course.C = SC.C
group by  Course.C , Course.Cname
order by Course.C , Course.Cname

–40、查询选修”张三”老师所授课程的学习者中,成绩最高的学子新闻及其成绩
–40.1 当最高分独有二个时
select top 1 Student.* , Course.Cname , SC.C , SC.score  
from Student, SC , Course , Teacher
where Student.S = SC.S and SC.C = Course.C and Course.T = Teacher.T and
Teacher.Tname = N’张三’
order by SC.score desc
–40.2 当最高分现身多个时
select Student.* , Course.Cname , SC.C , SC.score  
from Student, SC , Course , Teacher
where Student.S = SC.S and SC.C = Course.C and Course.T = Teacher.T and
Teacher.Tname = N’张三’ and
SC.score = (select max(SC.score) from SC , Course , Teacher where SC.C =
Course.C and Course.T = Teacher.T and Teacher.Tname = N’张三’)

–41、查询差异科目战表同样的学员的上学的小孩子编号、课程编号、学子战表 
–方法1
select m.* from SC m ,(select C , score from SC group by C , score
having count(1) > 1) n 
where m.C= n.C and m.score = n.score order by m.C , m.score , m.S
–方法2
select m.* from SC m where exists (select 1 from (select C , score from
SC group by C , score having count(1) > 1) n 
where m.C= n.C and m.score = n.score) order by m.C , m.score , m.S

–42、查询每门功战绩最棒的前两名 
select t.* from sc t where score in (select top 2 score from sc where C
= T.C order by score desc) order by t.C , t.score desc

–43、总括每门课程的学童选修人数(抢先5人的教程才总括)。要求输出课程号和选修人数,查询结果按人头降序排列,若人数相仿,按学科号升序排列  
select Course.C , Course.Cname , count(*) [学子人数]
from Course , SC 
where Course.C = SC.C
group by  Course.C , Course.Cname
having count(*) >= 5
order by [学子人数] desc , Course.C

–44、检索起码选修两门科目标学员学号 
select student.S , student.Sname 
from student , SC 
where student.S = SC.S 
group by student.S , student.Sname 
having count(1) >= 2
order by student.S

–45、查询选修了全部科目标学子消息 
–方法1 基于数据来成功
select student.* from student where S in
(select S from sc group by S having count(1) = (select count(1) from
course))
–方法2 使用重复否定来成功
select t.* from student t where t.S not in 
(
  select distinct m.S from
  (
    select S , C from student , course 
  ) m where not exists (select 1 from sc n where n.S = m.S and n.C =
m.C)
)
–方法3 接纳重复否定来成功
select t.* from student t where not exists(select 1 from 
(
  select distinct m.S from
  (
    select S , C from student , course 
  ) m where not exists (select 1 from sc n where n.S = m.S and n.C =
m.C)
) k where k.S = t.S
)

–46、查询各学子的年华
–46.1 只根据年度来算
select * , datediff(yy , sage , getdate()) [年龄] from student
–46.2 依据出生辰期来算,当前月日 < 出生年月的月日则,岁数减一
select * , case when right(convert(varchar(10),getdate(),120),5) <
right(convert(varchar(10),sage,120),5) then datediff(yy , sage ,
getdate()) – 1 else datediff(yy , sage , getdate()) end [年龄] from
student

–47、查询本周过生日的学员
select * from student where datediff(week,datename(yy,getdate()) +
right(convert(varchar(10),sage,120),6),getdate()) = 0

–48、查询前一周过寿辰的学习者
select * from student where datediff(week,datename(yy,getdate()) +
right(convert(varchar(10),sage,120),6),getdate()) = -1

–49、查询上一个月过华诞的上学的儿童
select * from student where datediff(mm,datename(yy,getdate()) +
right(convert(varchar(10),sage,120),6),getdate()) = 0

–50、查询上个月过生日的学员
select * from student where datediff(mm,datename(yy,getdate()) +
right(convert(varchar(10),sage,120),6),getdate()) = -1

drop table  Student,Course,Teacher,SC

 

–1、查询”01″课程比”02″课程成绩高的学员的新闻及教程分数
–1.1、查询同一时间存在”01″课程和”02″课程的情况
select a.* , b.score [课程’01’的分数],c.score [课程’02’的分数] from Student a , SC b , SC c
where a.S# = b.S# and a.S# = c.S# and b.C# = ’01’ and c.C# = ’02’ and b.score > c.score
–1.2、查询同期设有”01″课程和”02″课程的状态和存在”01″课程但恐怕不设有”02″课程的情景(一纸空文时呈现为null卡塔尔(قطر‎(以下存在一样内容时不再解释卡塔尔(قطر‎
select a.* , b.score [课程”01″的分数],c.score [课程”02″的分数] from Student a
left join SC b on a.S# = b.S# and b.C# = ’01’
left join SC c on a.S# = c.S# and c.C# = ’02’
where b.score > isnull(c.score,0)

/*数据表构造

–2、查询”01″课程比”02″课程战绩低的学童的新闻及学科分数
–2.1、查询同一时候设有”01″课程和”02″课程的状态
select a.* , b.score [课程’01’的分数],c.score [课程’02’的分数] from Student a , SC b , SC c
where a.S# = b.S# and a.S# = c.S# and b.C# = ’01’ and c.C# = ’02’ and b.score < c.score
–2.2、查询同期设有”01″课程和”02″课程的动静和空中楼阁”01″课程但存在”02″课程的气象
select a.* , b.score [课程”01″的分数],c.score [课程”02″的分数] from Student a
left join SC b on a.S# = b.S# and b.C# = ’01’
left join SC c on a.S# = c.S# and c.C# = ’02’
where isnull(b.score,0) < c.score

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