SQL练习题

数据表介绍

--1.学生表
Student(SId,Sname,Sage,Ssex)
--SId 学生编号,Sname 学生姓名,Sage 出生年月,Ssex 学生性别

--2.课程表
Course(CId,Cname,TId)
--CId 课程编号,Cname 课程名称,TId 教师编号

--3.教师表
Teacher(TId,Tname)
--TId 教师编号,Tname 教师姓名

--4.成绩表
SC(SId,CId,score)
--SId 学生编号,CId 课程编号,score 分数

学生表 Student

create table Student(SId varchar(10),Sname varchar(10),Sage datetime,Ssex varchar(10));

insert into Student values('01' , '赵雷' , '1990-01-01' , '男');

insert into Student values('02' , '钱电' , '1990-12-21' , '男');

insert into Student values('03' , '孙风' , '1990-12-20' , '男');

insert into Student values('04' , '李云' , '1990-12-06' , '男');

insert into Student values('05' , '周梅' , '1991-12-01' , '女');

insert into Student values('06' , '吴兰' , '1992-01-01' , '女');

insert into Student values('07' , '郑竹' , '1989-01-01' , '女');

insert into Student values('09' , '张三' , '2017-12-20' , '女');

insert into Student values('10' , '李四' , '2017-12-25' , '女');

insert into Student values('11' , '李四' , '2012-06-06' , '女');

insert into Student values('12' , '赵六' , '2013-06-13' , '女');

insert into Student values('13' , '孙七' , '2014-06-01' , '女');

科目表 Course

create table Course(CId varchar(10),Cname nvarchar(10),TId varchar(10));

insert into Course values('01' , '语文' , '02');

insert into Course values('02' , '数学' , '01');

insert into Course values('03' , '英语' , '03');

教师表 Teacher

create table Teacher(TId varchar(10),Tname varchar(10));

insert into Teacher values('01' , '张三');

insert into Teacher values('02' , '李四');

insert into Teacher values('03' , '王五');

成绩表 SC

create table SC(SId varchar(10),CId 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);

练习题目

1. 查询" 01 "课程比" 02 "课程成绩高的学生的信息及课程分数

因为需要全部的学生信息，则需要在sc表中得到符合条件的SId后与student表进行join，可以left join 也可以 right join

select * from Student RIGHT JOIN (
    select t1.SId, class1, class2 from
          (select SId, score as class1 from sc where sc.CId = '01')as t1, 
          (select SId, score as class2 from sc where sc.CId = '02')as t2
    where t1.SId = t2.SId AND t1.class1 > t2.class2
)r 
on Student.SId = r.SId;
select * from  (
    select t1.SId, class1, class2 
    from
        (SELECT SId, score as class1 FROM sc WHERE sc.CId = '01') AS t1, 
        (SELECT SId, score as class2 FROM sc WHERE sc.CId = '02') AS t2
    where t1.SId = t2.SId and t1.class1 > t2.class2
) r 
LEFT JOIN Student
ON Student.SId = r.SId;

1.1 查询同时存在" 01 "课程和" 02 "课程的情况

select * from 
    (select * from sc where sc.CId = '01') as t1, 
    (select * from sc where sc.CId = '02') as t2
where t1.SId = t2.SId;

1.2 查询存在" 01 "课程但可能不存在" 02 "课程的情况(不存在时显示为 null )

这一道就是明显需要使用join的情况了，02可能不存在，即为left join的右侧或right join 的左侧即可.

select * from 
(select * from sc where sc.CId = '01') as t1
left join 
(select * from sc where sc.CId = '02') as t2
on t1.SId = t2.SId;
select * from 
(select * from sc where sc.CId = '02') as t2
right join 
(select * from sc where sc.CId = '01') as t1
on t1.SId = t2.SId;

1.3 查询不存在" 01 "课程但存在" 02 "课程的情况

select * from sc
where sc.SId not in (
    select SId from sc 
    where sc.CId = '01'
) 
AND sc.CId= '02';

2. 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

这里只用根据学生ID把成绩分组，对分组中的score求平均值，最后在选取结果中AVG大于60的即可. 注意，这里必须要给计算得到的AVG结果一个alias.（AS ss）得到学生信息的时候既可以用join也可以用一般的联合搜索

select student.SId,sname,ss from student,(
    select SId, AVG(score) as ss from sc  
    GROUP BY SId 
    HAVING AVG(score)> 60
    )r
where student.sid = r.sid;
select Student.SId, Student.Sname, r.ss from Student right join(
      select SId, AVG(score) AS ss from sc
      GROUP BY SId
      HAVING AVG(score)> 60
)r on Student.SId = r.SId;
select s.SId,ss,Sname from(
select SId, AVG(score) as ss from sc  
GROUP BY SId
HAVING AVG(score)> 60
)r left join 
(select Student.SId, Student.Sname from
Student)s on s.SId = r.SId;

1. 3. 查询在 SC 表存在成绩的学生信息

2. select DISTINCT student.*
   from student,sc
   where student.SId=sc.SId

3. 4.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(没成绩的显示为 null )

4. 联合查询不会显示没选课的学生：

5. select student.sid, student.sname,r.coursenumber,r.scoresum
   from student,
   (select sc.sid, sum(sc.score) as scoresum, count(sc.cid) as coursenumber from sc 
   group by sc.sid)r
   where student.sid = r.sid;

   如要显示没选课的学生(显示为NULL)，需要使用join:

6. select s.sid, s.sname,r.coursenumber,r.scoresum
   from (
       (select student.sid,student.sname 
       from student
       )s 
       left join 
       (select 
           sc.sid, sum(sc.score) as scoresum, count(sc.cid) as coursenumber
           from sc 
           group by sc.sid
       )r 
      on s.sid = r.sid
   );

7. 4.1 查有成绩的学生信息

8. 这一题涉及到in和exists的用法，在这种小表中，两种方法的效率都差不多，但是请参考SQL查询中in和exists的区别分析
   当表2的记录数量非常大的时候，选用exists比in要高效很多.
   EXISTS用于检查子查询是否至少会返回一行数据，该子查询实际上并不返回任何数据，而是返回值True或False.
   结论：IN()适合B表比A表数据小的情况
   结论：EXISTS()适合B表比A表数据大的情况

9. select * from student 
   where exists (select sc.sid from sc where student.sid = sc.sid);
   select * from student
   where student.sid in (select sc.sid from sc);

   5. 查询「李」姓老师的数量

10. select count(*)
    from teacher
    where tname like '李%';

11. 6. 查询学过「张三」老师授课的同学的信息

12. 多表联合查询

13. select student.* from student,teacher,course,sc
    where 
        student.sid = sc.sid 
        and course.cid=sc.cid 
        and course.tid = teacher.tid 
        and tname = '张三';

14. 7. 查询没有学全所有课程的同学的信息

15. 因为有学生什么课都没有选，反向思考，先查询选了所有课的学生，再选择这些人之外的学生.

16. select * from student
    where student.sid not in (
      select sc.sid from sc
      group by sc.sid
      having count(sc.cid)= (select count(cid) from course)
    );

17. 8. 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

18. 这个用联合查询也可以，但是逻辑不清楚，我觉得较为清楚的逻辑是这样的：从sc表查询01同学的所有选课cid--从sc表查询所有同学的sid如果其cid在前面的结果中--从student表查询所有学生信息如果sid在前面的结果中

19. select * from student 
    where student.sid in (
        select sc.sid from sc 
        where sc.cid in(
            select sc.cid from sc 
            where sc.sid = '01'
        )
    );

20. 9. 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

21. 10. 查询没学过"张三"老师讲授的任一门课程的学生姓名

22. 仍然还是嵌套，三层嵌套， 或者多表联合查询

23. select * from student
        where student.sid not in(
            select sc.sid from sc where sc.cid in(
                select course.cid from course where course.tid in(
                    select teacher.tid from teacher where tname = "张三"
                )
            )
        );
    select * from student
    where student.sid not in(
        select sc.sid from sc,course,teacher 
        where
            sc.cid = course.cid
            and course.tid = teacher.tid
            and teacher.tname= "张三"
    );

24. 11. 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

25. 从SC表中选取score小于60的，并group by sid，having count 大于1

26. select student.SId, student.Sname,b.avg
    from student RIGHT JOIN
    (select sid, AVG(score) as avg from sc
        where sid in (
                  select sid from sc 
                  where score<60 
                  GROUP BY sid 
                  HAVING count(score)>1)
        GROUP BY sid) b on student.sid=b.sid;

27. 12. 检索" 01 "课程分数小于 60，按分数降序排列的学生信息

28. 双表联合查询，在查询最后可以设置排序方式，语法为ORDER BY ***** DESC\ASC;

29. select student.*, sc.score from student, sc
    where student.sid = sc.sid
    and sc.score < 60
    and cid = "01"
    ORDER BY sc.score DESC;

30. 13. 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

31. select *  from sc 
    left join (
        select sid,avg(score) as avscore from sc 
        group by sid
        )r 
    on sc.sid = r.sid
    order by avscore desc;

32. 14. 查询各科成绩最高分、最低分和平均分：

33. 以如下形式显示：课程 ID，课程 name，最高分，最低分，平均分，及格率，中等率，优良率，优秀率

    及格为>=60，中等为：70-80，优良为：80-90，优秀为：>=90

    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

34. select 
    sc.CId ,
    max(sc.score)as 最高分,
    min(sc.score)as 最低分,
    AVG(sc.score)as 平均分,
    count(*)as 选修人数,
    sum(case when sc.score>=60 then 1 else 0 end )/count(*)as 及格率,
    sum(case when sc.score>=70 and sc.score<80 then 1 else 0 end )/count(*)as 中等率,
    sum(case when sc.score>=80 and sc.score<90 then 1 else 0 end )/count(*)as 优良率,
    sum(case when sc.score>=90 then 1 else 0 end )/count(*)as 优秀率 
    from sc
    GROUP BY sc.CId
    ORDER BY count(*)DESC, sc.CId ASC

    15. 按各科成绩进行排序，并显示排名， Score 重复时保留名次空缺

35. 这一道题可以用变量，但也有更为简单的方法，即自交（左交）
    用sc中的score和自己进行对比，来计算“比当前分数高的分数有几个”。

36. select a.cid, a.sid, a.score, count(b.score)+1 as rank
    from sc as a 
    left join sc as b 
    on a.score

37. 15.1 按各科成绩进行排序，并显示排名， Score 重复时合并名次

38. 16. 查询学生的总成绩，并进行排名，总分重复时保留名次空缺

39. 16.1 查询学生的总成绩，并进行排名，总分重复时不保留名次空缺

40. 这里主要学习一下使用变量。在SQL里面变量用@来标识

41. set @crank=0;
    select q.sid, total, @crank := @crank +1 as rank from(
    select sc.sid, sum(sc.score) as total from sc
    group by sc.sid
    order by total desc)q;

42. 17. 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]，[60-0] 及所占百分比

43. group by以后的查询结果无法使用别名，所以不要想着先单表group by计算出结果再从第二张表里添上课程信息，而应该先将两张表join在一起得到所有想要的属性再对这张总表进行统计计算。这里就不算百分比了，道理相同。

44. 注意一下，用case when 返回1 以后的统计不是用count而是sum

45. select course.cname, course.cid,
    sum(case when sc.score<=100 and sc.score>85 then 1 else 0 end) as "[100-85]",
    sum(case when sc.score<=85 and sc.score>70 then 1 else 0 end) as "[85-70]",
    sum(case when sc.score<=70 and sc.score>60 then 1 else 0 end) as "[70-60]",
    sum(case when sc.score<=60 and sc.score>0 then 1 else 0 end) as "[60-0]"
    from sc left join course
    on sc.cid = course.cid
    group by sc.cid;

46. 18. 查询各科成绩前三名的记录

47. mysql不能group by 了以后取limit，所以不要想着讨巧了，我快被这一题气死了。思路有两种，第一种比较暴力，计算比自己分数大的记录有几条，如果小于3 就select，因为对前三名来说不会有3个及以上的分数比自己大了，最后再对所有select到的结果按照分数和课程编号排名即可。

48. select * from sc
    where (
    select count(*) from sc as a 
    where sc.cid = a.cid and sc.score

49. 19. 查询每门课程被选修的学生数

50. select cid, count(sid) from sc 
    group by cid;

51. 20. 查询出只选修两门课程的学生学号和姓名

52. 嵌套查询

53. select student.sid, student.sname from student
    where student.sid in
    (select sc.sid from sc
    group by sc.sid
    having count(sc.cid)=2
    );

    联合查询

54. select student.SId,student.Sname
    from sc,student
    where student.SId=sc.SId  
    GROUP BY sc.SId
    HAVING count(*)=2；

55. 21. 查询男生、女生人数

56. select ssex, count(*) from student
    group by ssex;

57. 22. 查询名字中含有「风」字的学生信息

58. select *
    from student 
    where student.Sname like '%风%'

59. 23. 查询同名同性学生名单，并统计同名人数

60. 找到同名的名字并统计个数

61. select sname, count(*) from student
    group by sname
    having count(*)>1;

    嵌套查询列出同名的全部学生的信息

62. select * from student
    where sname in (
    select sname from student
    group by sname
    having count(*)>1
    );

63. 24. 查询 1990 年出生的学生名单

64. select *
    from student
    where YEAR(student.Sage)=1990;

65. 25. 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

66. select sc.cid, course.cname, AVG(SC.SCORE) as average from sc, course
    where sc.cid = course.cid
    group by sc.cid 
    order by average desc,cid asc;

67. 26. 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

68. having也可以用来截取结果表，在这里就先得到平均成绩总表，再截取AVG大于85的即可.

69. select student.sid, student.sname, AVG(sc.score) as aver from student, sc
    where student.sid = sc.sid
    group by sc.sid
    having aver > 85;

70. 27. 查询课程名称为「数学」，且分数低于 60 的学生姓名和分数

71. select student.sname, sc.score from student, sc, course
    where student.sid = sc.sid
    and course.cid = sc.cid
    and course.cname = "数学"
    and sc.score < 60;

72. 28. 查询所有学生的课程及分数情况（存在学生没成绩，没选课的情况）

73. select student.sname, cid, score from student
    left join sc
    on student.sid = sc.sid;

74. 29. 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

75. select student.sname, course.cname,sc.score from student,course,sc
    where sc.score>70
    and student.sid = sc.sid
    and sc.cid = course.cid;

76. 30. 查询不及格的课程

77. 可以用group by 来取唯一，也可以用distinct

78. select cid from sc
    where score< 60
    group by cid;
    select DISTINCT sc.CId
    from sc
    where sc.score <60;

79. 31. 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

80. select student.sid,student.sname 
    from student,sc
    where cid="01"
    and score>=80
    and student.sid = sc.sid;

81. 32. 求每门课程的学生人数

82. select sc.CId,count(*) as 学生人数
    from sc
    GROUP BY sc.CId;

83. 33. 成绩不重复，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

84. 用having max()理论上也是对的，但是下面那种按分数排序然后取limit 1的更直观可靠

85. select student.*, sc.score, sc.cid from student, teacher, course,sc 
    where teacher.tid = course.tid
    and sc.sid = student.sid
    and sc.cid = course.cid
    and teacher.tname = "张三"
    having max(sc.score);
    select student.*, sc.score, sc.cid from student, teacher, course,sc 
    where teacher.tid = course.tid
    and sc.sid = student.sid
    and sc.cid = course.cid
    and teacher.tname = "张三"
    order by score desc
    limit 1;

86. 34. 成绩有重复的情况下，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

87. 为了验证这一题，先修改原始数据

88. UPDATE sc SET score=90
    where sid = "07"
    and cid ="02";

    这样张三老师教的02号课就有两个学生同时获得90的最高分了。
    这道题的思路继续上一题，我们已经查询到了符合限定条件的最高分了，这个时候只用比较这张表，找到全部score等于这个最高分的记录就可，看起来有点繁复。

89. select student.*, sc.score, sc.cid from student, teacher, course,sc 
    where teacher.tid = course.tid
    and sc.sid = student.sid
    and sc.cid = course.cid
    and teacher.tname = "张三"
    and sc.score = (
        select Max(sc.score) 
        from sc,student, teacher, course
        where teacher.tid = course.tid
        and sc.sid = student.sid
        and sc.cid = course.cid
        and teacher.tname = "张三"
    );

90. 35. 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

91. 同上，在这里用了inner join后会有概念是重复的记录：“01 课与 03课”=“03 课与 01 课”，所以这里取唯一可以直接用group by

92. select  a.cid, a.sid,  a.score from sc as a
    inner join 
    sc as b
    on a.sid = b.sid
    and a.cid != b.cid
    and a.score = b.score
    group by cid, sid;

93. 36. 查询每门功成绩最好的前两名

94. select a.sid,a.cid,a.score from sc as a 
    left join sc as b 
    on a.cid = b.cid and a.score

95. 37. 统计每门课程的学生选修人数（超过 5 人的课程才统计）。

96. select sc.cid, count(sid) as cc from sc
    group by cid
    having cc >5;

97. 38. 检索至少选修两门课程的学生学号

98. select sid, count(cid) as cc from sc
    group by sid
    having cc>=2;

99. 39. 查询选修了全部课程的学生信息

100. select student.*
     from sc ,student 
     where sc.SId=student.SId
     GROUP BY sc.SId
     HAVING count(*) = (select DISTINCT count(*) from course )

101. 40. 查询各学生的年龄，只按年份来算

102. 41. 按照出生日期来算，当前月日 < 出生年月的月日则，年龄减一

103. select student.SId as 学生编号,student.Sname  as  学生姓名,
     TIMESTAMPDIFF(YEAR,student.Sage,CURDATE()) as 学生年龄
     from student

104. 42. 查询本周过生日的学生

105. select *
     from student 
     where WEEKOFYEAR(student.Sage)=WEEKOFYEAR(CURDATE());

106. 43. 查询下周过生日的学生

107. select *
     from student 
     where WEEKOFYEAR(student.Sage)=WEEKOFYEAR(CURDATE())+1;

108. 44. 查询本月过生日的学生

109. select *
     from student 
     where MONTH(student.Sage)=MONTH(CURDATE());

110. 45. 查询下月过生日的学生

111. select *
     from student 
     where MONTH(student.Sage)=MONTH(CURDATE())+1;