sql-经典例子练习

 

/************************************************************************************************************

* 数据见50条常用sql.sql脚本

* 1.学生表

* student(studentID,stu_name,stu_age,stu_sex) --studentID 学生编号,stu_name 学生姓名,stu_age 出生年月,stu_sex 学生性别

* 2.课程表 

* course(courseID,crs_name,teacherID) --courseID --课程编号,crs_name 课程名称,teacherID 教师编号

* 3.教师表 

* teacher(teacherID,tea_name) --teacherID 教师编号,tea_name 教师姓名

* 4.成绩表 

* stu_crs(studentID,courseID,score) --studentID 学生编号,courseID 课程编号,score 分数

        ////"01"课程、"02"课程是变量（记录）。不是常量（字段）。

        ////如果把"01"、"02"设计成字段，就会使难度降低，但是不够灵活。比如以后会再加一科目04物理。

 

* ########方法：先确定哪些表的字段，再一一满足条件，即可。

************************************************************************************************************/

表达式：where isnull(b.score,0)是简单的动态，还有动态sql。

连接顺序：选择那个表作为源数据表，那个表优先被连接

3、group by子句将数据划分为多个分组； 

5、使用having子句筛选分组； 

7、使用order by对结果集进行排序。

 

group by 、左外连接和where并用。

--1、查询"01"课程比"02"课程成绩高的学生的信息及课程分数        

        #####同一个表同一个字段值的比较，把一个表变为两个表，才能比较。

        #####正常的是，一个字段值和一个常量比较。

        有两种情况：1.1、1.2

--1.1、同时存在"01"课程和"02"课程的情况(01必选课，02必选课)

        SELECT  a.*,b.score,c.score 

        FROM student a,stu_crs b,stu_crs c

        WHERE b.courseID='01' AND c.courseID='02' AND b.score>c.score AND a.studentID=b.studentID AND  c.studentID=a.studentID

        (a.studentID=b.studentID AND  c.studentID=a.studentID这两个都需要，因为c.studentID和b.studentID与a表关联的结果集不一样)

--1.2、存在"01"课程但可能不存在"02"课程的情况、(01必选课，02选修课)、(不存在时显示为null)(以下存在相同内容时不再解释)

        select * from student a 

        left join stu_crs b on a.studentID=b.studentID and b.courseID='01'

        left join stu_crs c on a.studentID=c.studentID and c.courseID='04'

        and b.score>c.score (and 有8条，where b.score>c.score只有一条）

 

 

--3、查询平均成绩大于等于60分的同学的学生编号和学生姓名和平均成绩

        select a.*,avg(b.score) from student a,stu_crs b

        where a.studentID = b.studentID

        group by a.studentID

        having avg(b.score)>70

        order by avg(b.score) desc

 

--5、查询所有同学的学生编号、学生姓名、选课总数、所有课程的总成绩

        --5.1、查询所有有成绩的SQL。

        select a.*,count(b.score),sum(b.score) from student a,stu_crs b

        where a.studentID = b.studentID                                        #######因为“表达式需要b.score”，且b.score可能有空的。resultSet7条。

        group by a.studentID

        --5.2、查询所有(包括有成绩和无成绩)的SQL。

        select a.*,count(b.score),sum(b.score) from student a

        left join stu_crs b on a.studentID = b.studentID                        #######left join，因为b.score可能有空的。resultSet8条。

        group by a.studentID                                                                #######对结果集还要进行计算的，一般需要group by、having

 

--11、查询没有学全所有课程的同学的信息 

select * from student a,stu_crs b,course c

where a.studentID=b.studentID

group by a.studentID

having count(distinct b.courseID)         ######不能是count(distinct b.courseID)==1，因为group by a.studentID过了，

< count(distinct c.courseID)

 

--12、查询至少有一门课与学号为"01"的同学所学相同的同学的信息 

select distinct                                         ########distinct过滤重复结果

a.* from student a, stu_crs b where a.studentID = b.studentID and b.courseID 

in (select courseID from stu_crs where studentID = '01') and a.studentID <> '01'

 

--13、查询和"01"号的同学学习的课程完全相同的其他同学的信息 

select student.* from student where studentID in

(select distinct stu_crs.studentID from stu_crs where studentID <> '01' 

and stu_crs.courseID in (select distinct courseID from stu_crs where studentID = '01')         ######限定范围和限定数量一块就可以确定完全一样了。

group by stu_crs.studentID having count(1) = (select count(1) from stu_crs where studentID='01')) ###限定数量，count(1) 与count(*)效果一样。

 

--15、查询两门及其以上不及格课程的同学的学号，姓名及其平均成绩 

        select a.*,avg(b.score) from student a,stu_crs b 

        where a.studentID=b.studentID and a.studentID 

        in(select studentID from stu_crs where score <60 group by studentID having count(1)>1)

        group by a.studentID

 

 

 

 

 

 

 

--------------------------------未完成--------------------------------

 

--17、按平均成绩从高到低显示所有学生的所有课程的成绩以及平均成绩

select a.studentID  , a.stu_name ,avg(b.score) ps

       max(case c.crs_name when N'语文' then b.score else null end) [语文],        #####表达式？？？？？？？？？？？

from student a 

left join stu_crs b on a.studentID = b.studentID

left join course c on b.courseID = c.courseID

group by a.studentID , a.stu_name

order by ps desc                                                                #####order by时可以用别名，不用再查一次了。

 

 

--21、查询不同老师所教不同课程平均分从高到低显示 

select m.teacherID , m.tea_name , cast(avg(o.score) as decimal(18,2)) avg_score

from teacher m , course n , stu_crs o

where m.teacherID = n.teacherID and n.courseID = o.courseID

group by m.teacherID , m.tea_name

order by avg_score destu_crs

 

--22、查询所有课程的成绩第2名到第3名的学生信息及该课程成绩

--22.1 sql 2000用子查询完成

--score重复时保留名次空缺

select * from (select t.* , px = (select count(1) from stu_crs where courseID = t.courseID and score > t.score) + 1 from stu_crs t) m where px between 2 and 3 order by m.courseID , m.px 

--score重复时合并名次

select * from (select t.* , px = (select count(distinct score) from stu_crs where courseID = t.courseID and score >= t.score) from stu_crs t) m where px between 2 and 3 order by m.courseID , m.px 

--22.2 sql 2005用rank,DENSE_RANK完成

--score重复时保留名次空缺(rank完成)

select * from (select t.* , px = rank() over(partition by courseID order by score destu_crs) from stu_crs t) m where px between 2 and 3 order by m.courseID , m.px 

--score重复时合并名次(DENSE_RANK完成)

select * from (select t.* , px = DENSE_RANK() over(partition by courseID order by score destu_crs) from stu_crs t) m where px between 2 and 3 order by m.courseID , m.px 

 

--23、统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],[0-60]及所占百分比 

--23.1 统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],[0-60]

--横向显示

select course.courseID [课程编号] , crs_name 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 stu_crs , course 

where stu_crs.courseID = course.courseID 

group by course.courseID , course.crs_name

order by course.courseID

--纵向显示1(显示存在的分数段)

select m.courseID [课程编号] , m.crs_name [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 , stu_crs n

where m.courseID = n.courseID 

group by m.courseID , m.crs_name , (

  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.courseID , m.crs_name , 分数段

--纵向显示2(显示存在的分数段，不存在的分数段用0显示)

select m.courseID [课程编号] , m.crs_name [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 , stu_crs n

where m.courseID = n.courseID 

group by all m.courseID , m.crs_name , (

  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.courseID , m.crs_name , 分数段

 

--23.2 统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 

--横向显示

select m.courseID 课程编号, m.crs_name 课程名称,

  (select count(1) from stu_crs where courseID = m.courseID and score < 60) [0-60],

  cast((select count(1) from stu_crs where courseID = m.courseID and score < 60)*100.0 / (select count(1) from stu_crs where courseID = m.courseID) as decimal(18,2)) [百分比(%)],

  (select count(1) from stu_crs where courseID = m.courseID and score >= 60 and score < 70) [60-70],

  cast((select count(1) from stu_crs where courseID = m.courseID and score >= 60 and score < 70)*100.0 / (select count(1) from stu_crs where courseID = m.courseID) as decimal(18,2)) [百分比(%)],

  (select count(1) from stu_crs where courseID = m.courseID and score >= 70 and score < 85) [70-85],

  cast((select count(1) from stu_crs where courseID = m.courseID and score >= 70 and score < 85)*100.0 / (select count(1) from stu_crs where courseID = m.courseID) as decimal(18,2)) [百分比(%)],

  (select count(1) from stu_crs where courseID = m.courseID and score >= 85) [85-100],

  cast((select count(1) from stu_crs where courseID = m.courseID and score >= 85)*100.0 / (select count(1) from stu_crs where courseID = m.courseID) as decimal(18,2)) [百分比(%)]

from course m 

order by m.courseID

--纵向显示1(显示存在的分数段)

select m.courseID [课程编号] , m.crs_name [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 stu_crs where courseID = m.courseID) as decimal(18,2)) [百分比(%)]

from course m , stu_crs n

where m.courseID = n.courseID 

group by m.courseID , m.crs_name , (

  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.courseID , m.crs_name , 分数段

--纵向显示2(显示存在的分数段，不存在的分数段用0显示)

select m.courseID [课程编号] , m.crs_name [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 stu_crs where courseID = m.courseID) as decimal(18,2)) [百分比(%)]

from course m , stu_crs n

where m.courseID = n.courseID 

group by all m.courseID , m.crs_name , (

  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.courseID , m.crs_name , 分数段

 

--24、查询学生平均成绩及其名次 

--24.1 查询学生的平均成绩并进行排名，sql 2000用子查询完成，分平均成绩重复时保留名次空缺和不保留名次空缺两种。

select t1.* , px = (select count(1) from 

(

  select m.studentID [学生编号] , 

         m.stu_name [学生姓名] ,

         isnull(cast(avg(score) as decimal(18,2)),0) [平均成绩]

  from student m left join stu_crs n on m.studentID = n.studentID 

  group by m.studentID , m.stu_name

) t2 where 平均成绩 > t1.平均成绩) + 1 from 

(

  select m.studentID [学生编号] , 

         m.stu_name [学生姓名] ,

         isnull(cast(avg(score) as decimal(18,2)),0) [平均成绩]

  from student m left join stu_crs n on m.studentID = n.studentID 

  group by m.studentID , m.stu_name

) t1

order by px

 

select t1.* , px = (select count(distinct 平均成绩) from 

(

  select m.studentID [学生编号] , 

         m.stu_name [学生姓名] ,

         isnull(cast(avg(score) as decimal(18,2)),0) [平均成绩]

  from student m left join stu_crs n on m.studentID = n.studentID 

  group by m.studentID , m.stu_name

) t2 where 平均成绩 >= t1.平均成绩) from 

(

  select m.studentID [学生编号] , 

         m.stu_name [学生姓名] ,

         isnull(cast(avg(score) as decimal(18,2)),0) [平均成绩]

  from student m left join stu_crs n on m.studentID = n.studentID 

  group by m.studentID , m.stu_name

) t1

order by px

--24.2 查询学生的平均成绩并进行排名，sql 2005用rank,DENSE_RANK完成，分平均成绩重复时保留名次空缺和不保留名次空缺两种。

select t.* , px = rank() over(order by [平均成绩] destu_crs) from

(

  select m.studentID [学生编号] , 

         m.stu_name [学生姓名] ,

         isnull(cast(avg(score) as decimal(18,2)),0) [平均成绩]

  from student m left join stu_crs n on m.studentID = n.studentID 

  group by m.studentID , m.stu_name

) t

order by px

 

select t.* , px = DENSE_RANK() over(order by [平均成绩] destu_crs) from

(

  select m.studentID [学生编号] , 

         m.stu_name [学生姓名] ,

         isnull(cast(avg(score) as decimal(18,2)),0) [平均成绩]

  from student m left join stu_crs n on m.studentID = n.studentID 

  group by m.studentID , m.stu_name

) t

order by px

 

--25、查询各科成绩前三名的记录

--25.1 分数重复时保留名次空缺

select m.* , n.courseID , n.score from student m, stu_crs n where m.studentID = n.studentID and n.score in 

(select top 3 score from stu_crs where courseID = n.courseID order by score destu_crs) order by n.courseID , n.score destu_crs

--25.2 分数重复时不保留名次空缺，合并名次

--sql 2000用子查询实现

select * from (select t.* , px = (select count(distinct score) from stu_crs where courseID = t.courseID and score >= t.score) from stu_crs t) m where px between 1 and 3 order by m.courseID , m.px 

--sql 2005用DENSE_RANK实现

select * from (select t.* , px = DENSE_RANK() over(partition by courseID order by score destu_crs) from stu_crs t) m where px between 1 and 3 order by m.courseID , m.px 

 

--26、查询每门课程被选修的学生数 

select courseID , count(studentID)[学生数] from stu_crs group by courseID

 

--27、查询出只有两门课程的全部学生的学号和姓名 

select student.studentID , student.stu_name

from student , stu_crs 

where student.studentID = stu_crs.studentID 

group by student.studentID , student.stu_name

having count(stu_crs.courseID) = 2

order by student.studentID

 

--28、查询男生、女生人数 

select count(stu_sex) as 男生人数 from student where stu_sex = N'男'

select count(stu_sex) as 女生人数 from student where stu_sex = N'女'

select sum(case when stu_sex = N'男' then 1 else 0 end) [男生人数],sum(case when stu_sex = N'女' then 1 else 0 end) [女生人数] from student

select case when stu_sex = N'男' then N'男生人数' else N'女生人数' end [男女情况] , count(1) [人数] from student group by case when stu_sex = N'男' then N'男生人数' else N'女生人数' end

 

--29、查询名字中含有"风"字的学生信息

select * from student where stu_name like N'%风%'

select * from student where charindex(N'风' , stu_name) > 0

 

--30、查询同名同性学生名单，并统计同名人数 

select stu_name [学生姓名], count(*) [人数] from student group by stu_name having count(*) > 1

 

--31、查询1990年出生的学生名单(注：student表中stu_age列的类型是datetime) 

select * from student where year(stu_age) = 1990

select * from student where datediff(yy,stu_age,'1990-01-01') = 0

select * from student where datepart(yy,stu_age) = 1990

select * from student where convert(varchar(4),stu_age,120) = '1990'

 

--32、查询每门课程的平均成绩，结果按平均成绩降序排列，平均成绩相同时，按课程编号升序排列 

select m.courseID , m.crs_name , cast(avg(n.score) as decimal(18,2)) avg_score

from course m, stu_crs n 

where m.courseID = n.courseID    

group by m.courseID , m.crs_name 

order by avg_score destu_crs, m.courseID astu_crs

 

--33、查询平均成绩大于等于85的所有学生的学号、姓名和平均成绩 

select a.studentID , a.stu_name , cast(avg(b.score) as decimal(18,2)) avg_score

from student a , stu_crs b

where a.studentID = b.studentID

group by a.studentID , a.stu_name

having cast(avg(b.score) as decimal(18,2)) >= 85 

order by a.studentID

 

--34、查询课程名称为"数学"，且分数低于60的学生姓名和分数 

select stu_name , score

from student , stu_crs , course 

where stu_crs.studentID = student.studentID and stu_crs.courseID = course.courseID and course.crs_name = N'数学' and score < 60 

 

--35、查询所有学生的课程及分数情况； 

select student.* , course.crs_name , stu_crs.courseID , stu_crs.score  

from student, stu_crs , course 

where student.studentID = stu_crs.studentID and stu_crs.courseID = course.courseID 

order by student.studentID , stu_crs.courseID

 

--36、查询任何一门课程成绩在70分以上的姓名、课程名称和分数； 

select student.* , course.crs_name , stu_crs.courseID , stu_crs.score  

from student, stu_crs , course 

where student.studentID = stu_crs.studentID and stu_crs.courseID = course.courseID and stu_crs.score >= 70 

order by student.studentID , stu_crs.courseID 

 

--37、查询不及格的课程

select student.* , course.crs_name , stu_crs.courseID , stu_crs.score  

from student, stu_crs , course 

where student.studentID = stu_crs.studentID and stu_crs.courseID = course.courseID and stu_crs.score < 60 

order by student.studentID , stu_crs.courseID 

 

--38、查询课程编号为01且课程成绩在80分以上的学生的学号和姓名； 

select student.* , course.crs_name , stu_crs.courseID , stu_crs.score  

from student, stu_crs , course 

where student.studentID = stu_crs.studentID and stu_crs.courseID = course.courseID and stu_crs.courseID = '01' and stu_crs.score >= 80 

order by student.studentID , stu_crs.courseID 

 

--39、求每门课程的学生人数 

select course.courseID , course.crs_name , count(*) [学生人数]

from course , stu_crs 

where course.courseID = stu_crs.courseID

group by  course.courseID , course.crs_name

order by course.courseID , course.crs_name

 

--40、查询选修"张三"老师所授课程的学生中，成绩最高的学生信息及其成绩

--40.1 当最高分只有一个时

select top 1 student.* , course.crs_name , stu_crs.courseID , stu_crs.score  

from student, stu_crs , course , teacher

where student.studentID = stu_crs.studentID and stu_crs.courseID = course.courseID and course.teacherID = teacher.teacherID and teacher.tea_name = N'张三'

order by stu_crs.score destu_crs

--40.2 当最高分出现多个时

select student.* , course.crs_name , stu_crs.courseID , stu_crs.score  

from student, stu_crs , course , teacher

where student.studentID = stu_crs.studentID and stu_crs.courseID = course.courseID and course.teacherID = teacher.teacherID and teacher.tea_name = N'张三' and

stu_crs.score = (select max(stu_crs.score) from stu_crs , course , teacher where stu_crs.courseID = course.courseID and course.teacherID = teacher.teacherID and teacher.tea_name = N'张三')

 

--41、查询不同课程成绩相同的学生的学生编号、课程编号、学生成绩 

--方法1

select m.* from stu_crs m ,(select courseID , score from stu_crs group by courseID , score having count(1) > 1) n 

where m.courseID= n.courseID and m.score = n.score order by m.courseID , m.score , m.studentID

--方法2

select m.* from stu_crs m where exists (select 1 from (select courseID , score from stu_crs group by courseID , score having count(1) > 1) n 

where m.courseID= n.courseID and m.score = n.score) order by m.courseID , m.score , m.studentID

 

--42、查询每门功成绩最好的前两名 

select t.* from stu_crs t where score in (select top 2 score from stu_crs where courseID = T.courseID order by score destu_crs) order by t.courseID , t.score destu_crs

 

--43、统计每门课程的学生选修人数（超过5人的课程才统计）。要求输出课程号和选修人数，查询结果按人数降序排列，若人数相同，按课程号升序排列  

select course.courseID , course.crs_name , count(*) [学生人数]

from course , stu_crs 

where course.courseID = stu_crs.courseID

group by  course.courseID , course.crs_name

having count(*) >= 5

order by [学生人数] destu_crs , course.courseID 

 

--44、检索至少选修两门课程的学生学号 

select student.studentID , student.stu_name 

from student , stu_crs 

where student.studentID = stu_crs.studentID 

group by student.studentID , student.stu_name 

having count(1) >= 2

order by student.studentID 

 

--45、查询选修了全部课程的学生信息 

--方法1 根据数量来完成

select student.* from student where studentID in

(select studentID from stu_crs group by studentID having count(1) = (select count(1) from course))

--方法2 使用双重否定来完成

select t.* from student t where t.studentID not in 

(

  select distinct m.studentID from

  (

    select studentID , courseID from student , course 

  ) m where not exists (select 1 from stu_crs n where n.studentID = m.studentID and n.courseID = m.courseID)

)

--方法3 使用双重否定来完成

select t.* from student t where not exists(select 1 from 

(

  select distinct m.studentID from

  (

    select studentID , courseID from student , course 

  ) m where not exists (select 1 from stu_crs n where n.studentID = m.studentID and n.courseID = m.courseID)

) k where k.studentID = t.studentID

)

 

--46、查询各学生的年龄

--46.1 只按照年份来算

select * , datediff(yy , stu_age , getdate()) [年龄] from student

--46.2 按照出生日期来算，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),stu_age,120),5) then datediff(yy , stu_age , getdate()) - 1 else datediff(yy , stu_age , getdate()) end [年龄] from student

 

--47、查询本周过生日的学生

select * from student where datediff(week,datename(yy,getdate()) + right(convert(varchar(10),stu_age,120),6),getdate()) = 0

 

--48、查询下周过生日的学生

select * from student where datediff(week,datename(yy,getdate()) + right(convert(varchar(10),stu_age,120),6),getdate()) = -1

 

--49、查询本月过生日的学生

select * from student where datediff(mm,datename(yy,getdate()) + right(convert(varchar(10),stu_age,120),6),getdate()) = 0

 

--50、查询下月过生日的学生

select * from student where datediff(mm,datename(yy,getdate()) + right(convert(varchar(10),stu_age,120),6),getdate()) = -1

 

drop table  student,course,teacher,stu_crs

评论

1 楼 zk302 2016-01-15  

格式可以优化下