50道SQL练习题及答案

作者: 一叶云秋 | 来源:发表于2019-06-07 22:12 被阅读47次

网上流传这这样的50道练习题，以下是我的解法，仅供各位联系和参考

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

join课程1成绩表和课程2成绩表，然后对该表进行筛选

SELECT
    c.SId,
    d.Sname,
    d.Sage,
    d.Ssex,
    c.S_01,
    c.S_02
FROM
    (
        SELECT
            a.SId,
            a.score AS S_01,
            b.score AS S_02
        FROM
            (
                SELECT
                    *
                FROM
                    50exercises.sc
                WHERE
                    CId = 01
            ) AS a
        JOIN (
            SELECT
                *
            FROM
                50exercises.sc
            WHERE
                CId = 02
        ) AS b ON a.SId = b.SId
        WHERE
            a.score > b.score
    ) AS c
JOIN 50exercises.student AS d ON c.SId = d.SId

1.1. 查询同时存在" 01 "课程和" 02 "课程的情况

解题思路 join课程1和课程2的两张表

SELECT
    a.SId,
    a.score AS S_01,
    b.score AS S_02
FROM
    (
        SELECT
            *
        FROM
            50exercises.sc
        WHERE
            CId = 01
    ) AS a
JOIN (
    SELECT
        *
    FROM
        50exercises.sc
    WHERE
        CId = 02
) AS b ON a.SId = b.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 )

left join课程1和课程2

SELECT
    a.SId,
    a.score AS S_01,
    b.score AS S_02
FROM
    (
        SELECT
            *
        FROM
            50exercises.sc
        WHERE
            CId = 01
    ) AS a
left JOIN (
    SELECT
        *
    FROM
        50exercises.sc
    WHERE
        CId = 02
) AS b ON a.SId = b.SId

1.3. 查询不存在" 01 "课程但存在" 02 "课程的情况

与1.2思路相似，反之即可

SELECT
    a.SId,
    a.score AS S_02,
    b.score AS S_01
FROM
    (
        SELECT
            *
        FROM
            50exercises.sc
        WHERE
            CId = 02
    ) AS a
LEFT JOIN (
    SELECT
        *
    FROM
        50exercises.sc
    WHERE
        CId = 01
) AS b ON a.SId = b.SId
where b.score IS NULL

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

筛选出平均成绩大于60分，和学生表Join

SELECT
    b.Sid,
    b.Sname,
    c.avg_s
FROM
    50exercises.student AS b
JOIN (
    SELECT
        a.SId,
        avg(a.score) AS avg_s
    FROM
        50exercises.sc AS a
    GROUP BY
        a.Sid
    HAVING
        avg(a.score) >= 60
) AS c ON b.SId = c.SId

3. 查询在 SC 表存在成绩的学生信息

找出SC表中的学生，然后join学生表

SELECT
    a.*
FROM
    50exercises.student AS a
JOIN (
    SELECT DISTINCT
        SId
    FROM
        50exercises.sc
) AS b ON a.SId = b.SId

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 )

从学生表left join成绩表

SELECT
    a.*, b.sum_s
FROM
    50exercises.student AS a
LEFT JOIN (
    SELECT
        SId,
        sum(score) AS sum_s
    FROM
        50exercises.sc
    GROUP BY
        SId
) AS b ON a.SId = b.SId

4.1 查有成绩的学生信息

从成绩表开始，然后匹配学生表即可

SELECT
    SId,
    sum(score) AS sum_s
FROM
    50exercises.sc
GROUP BY
    SId

5. 查询「李」姓老师的数量

找到姓李的老师的表，然后count

SELECT
    count(1)
FROM
    50exercises.teacher
WHERE Tname LIKE '李%'

6. 查询学过「张三」老师授课的同学的信息

先找出张三老师教的课，然后找到有这么课成绩的学生，然后join学生表

SELECT
    *
FROM
    50exercises.student AS e
JOIN (
    SELECT
        c.SId
    FROM
        50exercises.sc AS c
    JOIN (
        SELECT
            a.CId
        FROM
            50exercises.course AS a
        JOIN (
            SELECT
                *
            FROM
                50exercises.teacher
            WHERE
                Tname = '张三'
        ) AS b ON a.TId = b.TId
    ) AS d ON c.CId = d.CId
) AS f ON e.SId = f.SId

7. 查询没有学全所有课程的同学的信息

解法1，由于CId是整数，所以可以相加，那么只要找出不是6的或者为空的就可以

SELECT
    *
FROM
    student AS a
LEFT JOIN (
    SELECT
        SId,
        SUM(CId) AS sum_cid
    FROM
        sc
    GROUP BY
        SId
) AS b ON a.SId = b.SId
WHERE
    b.sum_cid <> 6
OR b.sum_cid IS NULL

标准解法，上过所有课的人，在sc表中出现的次数和课程表中课的数量相同，然后反选即可

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)
    )

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

首先得到学号01的同学学过的课程

select `CId`
from `50exercises`.sc
where `SId`= 01

然后取出至少就是筛选课程在这个范围内的记录，得到Sid

select distinct `SId`
from `50exercises`.sc
where `CId`in(
    select `CId`
    from `50exercises`.sc
    where `SId`= 01
    )

最后和信息表组合

select b.*
from
(
select distinct `SId`
from `50exercises`.sc
where `CId`in(
    select `CId`
    from `50exercises`.sc
    where `SId`= 01
    )
    ) as a
join `50exercises`.student as b
on a.SId = b.`SId`

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

select c.*
from `50exercises`.sc as a
join `50exercises`.student as c 
on a.`SId`=c.`SId`
group by a.SId
having  group_concat(a.CId order by a.CId) = (
        select  group_concat(b.CId order by b.CId)
        from `50exercises`.sc as b
        where b.SId = "01"
        group by b.SId
        )

10. 查询没学过"张三"老师讲授的任一门课程的学生姓名

select distinct d.`Sname`
from `50exercises`.teacher as a
join `50exercises`.course as b
on a.`TId`= b.`TId` and a.`Tname` != "张三"
join `50exercises`.sc as c 
on b.`CId` = c.`CId`
join `50exercises`.student as d
on c.`SId` =  d.`SId`

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

select a.`SId`,
            b.`Sname`,
            avg(a.score) as avg_score,
            sum(if(a.score < 60,1,0)) as smark
from `50exercises`.sc as a
join `50exercises`.student as b 
on a.`SId` = b.`SId`
group by a.`SId`
having smark >=2

12. 检索" 01 "课程分数小于 60，按分数降序排列的学生信息

SELECT
    *
FROM
    sc AS a
JOIN student AS b ON a.SId = b.SId
WHERE
    a.CId = "01"
AND a.score < 60
ORDER BY
    a.score DESC

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

SELECT
    *
FROM
    sc AS a
JOIN (
    SELECT
        Sid,
        avg(score) AS AVG_S
    FROM
        sc
    GROUP BY
        SId
) AS b ON a.SId = b.Sid
ORDER BY
    b.AVG_S DESC

14. 查询各科成绩最高分、最低分和平均分；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

SELECT
    b.Cname,
    b.CId,
    max(a.score) AS "最高分",
    min(a.score) AS "最低分",
    avg(a.score) AS "平均分",
    sum(

        IF (a.score >= 60 AND a.score < 70, 1, 0)
    ) / count(DISTINCT Sid) "及格率",
    sum(

        IF (a.score >= 70 AND a.score < 80, 1, 0)
    ) / count(DISTINCT Sid) "中等率",
    sum(

        IF (a.score >= 80 AND a.score < 90, 1, 0)
    ) / count(DISTINCT Sid) "优良率",
    sum(IF(a.score >= 90, 1, 0)) / count(DISTINCT Sid) "优秀率"
FROM
    sc AS a
JOIN course AS b ON a.CId = b.CId
GROUP BY
    CId

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

SELECT
    A.CId,
    A.SId,
    A.score,

IF (
    a.cid =@tmid ,@rank :=@rank + 1 ,@rank := 1
) AS rank ,@tmid := a.cid
FROM
    (SELECT * from SC order by cid asc,score DESC) AS A,
    (SELECT @rank := 0 ,@tmid := NULL) AS B

15.1 按各科成绩进行排序，并显示排名， Score 重复时合并名次

SELECT
    A.CId,
    A.SId,
    A.score,
    CASE
WHEN a.cid <> @tmid THEN
    @rank := 1
WHEN a.cid = @tmid
AND a.score <> @tscore THEN
    @rank :=@rank + 1
WHEN a.cid = @tmid
AND a.score = @tscore THEN
    @rank := @rank
ELSE @rank := 1
END AS rank,
 @tmid := a.cid,
 @tscore := a.score
FROM
    (
        SELECT
            *
        FROM
            SC
        ORDER BY
            cid ASC,
            score DESC
    ) AS A,
    (
        SELECT
            @rank := 1 ,@tmid := NULL ,@tscore := 0
    ) AS B

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

看了下数据，没有总成绩相同的学生，检验不出效果，所以改成用01课程来检验

SELECT
a.SId,
a.all_s,
CASE
WHEN a.all_s <> @tscore then @rank := @rank+1
ELSE @rank = @rank
END as rank,
@tscore := a.all_s
FROM
    (
        SELECT
            SId,
            sum(score) AS all_s
        FROM
            sc
        WHERE
            Cid = "01"
        GROUP BY
            SId
        ORDER BY
            all_s DESC
    ) AS a,
(SELECT @rank := 0 ,@tscore := 0) AS b

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

SELECT
a.SId,
a.all_s,
@rank := @rank+1 as rank
FROM
    (
        SELECT
            SId,
            sum(score) AS all_s
        FROM
            sc
        WHERE
            Cid = "01"
        GROUP BY
            SId
        ORDER BY
            all_s DESC
    ) AS a,
(SELECT @rank := 0 ) AS b

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

和14题相同

SELECT
    b.Cname,
    b.CId,
    sum(

        IF (a.score >= 0 AND a.score < 60, 1, 0)
    ) / count(DISTINCT Sid) "[0,60)",
    sum(

        IF (a.score >= 60 AND a.score < 70, 1, 0)
    ) / count(DISTINCT Sid) "[60,70)",
    sum(

        IF (a.score >= 70 AND a.score < 85, 1, 0)
    ) / count(DISTINCT Sid) "[70,85)",
    sum(IF(a.score >= 85, 1, 0)) / count(DISTINCT Sid) "[85,100]"
FROM
    sc AS a
JOIN course AS b ON a.CId = b.CId
GROUP BY
    CId

18. 查询各科成绩前三名的记录

SELECT
    *
FROM
    sc
WHERE
    (
        SELECT
            count(*)
        FROM
            sc AS a
        WHERE
            sc.cid = a.cid
        AND sc.score < a.score
    ) < 3 ##是不是比自己大的有三条
ORDER BY
    cid ASC,
    sc.score DESC;

解法2

SELECT
    a.sid,
    a.cid,
    a.score
FROM
    sc a
LEFT JOIN sc b ON a.cid = b.cid
AND a.score < b.score
GROUP BY
    a.cid,
    a.sid
HAVING
    count(b.cid) < 3
ORDER BY
    a.cid;

19. 查询每门课程被选修的学生数

SELECT
    CId,
    COUNT(DISTINCT SId) AS count_s
FROM
    SC
GROUP BY
    CId

20. 查询出只选修两门课程的学生学号和姓名

SELECT
    *
FROM
    (
        SELECT
            SId,
            COUNT(DISTINCT CId) AS count_C
        FROM
            SC
        GROUP BY
            SId
        HAVING
            count_C <= 2
    ) AS a
JOIN student AS b ON a.SId = b.SId

21. 查询男生、女生人数

SELECT
    Ssex,
    COUNT(DISTINCT SId) as Qty
FROM
    student
GROUP BY
    Ssex

22. 查询名字中含有「风」字的学生信息

SELECT
    *
FROM
    student
WHERE
    sname LIKE "%风%"

23. 查询同名同性学生名单，并统计同名人数

SELECT
    sname,
    COUNT(DISTINCT sid) AS Qty
FROM
    student
GROUP BY
    Sname
HAVING
    qty > 1

24. 查询 1990 年出生的学生名单

SELECT
    *
FROM
    student
WHERE
    YEAR (Sage) = 1990

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

SELECT
    CId,
    avg(score) AS avg_s
FROM
    sc
GROUP BY
    CId
ORDER BY
    avg_s DESC,
    CId ASC

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

SELECT
    a.SId,
    a.Sname,
    b.avg_s
FROM
    student AS a
JOIN (
    SELECT
        SId,
        avg(score) AS avg_s
    FROM
        sc
    GROUP BY
        SId
) AS b ON a.SId = b.SId

27. 查询课程名称为「数学」，且分数低于 60 的学生姓名和分数

SELECT
    c.Sname,
    d.score
FROM
    (
        SELECT
            a.SId,
            a.score
        FROM
            sc AS a
        JOIN course AS b ON a.CId = b.CId
        WHERE
            a.score < 60
        AND b.Cname = '数学'
    ) AS d
JOIN student AS c ON c.SId = d.SId

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）

SELECT
    *
FROM
    student AS a
JOIN course AS b
LEFT JOIN sc AS c ON a.SId = c.SId
AND b.CId = c.CId

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

SELECT
    *
FROM
    student AS a
JOIN course AS b
LEFT JOIN sc AS c ON a.SId = c.SId
AND b.CId = c.CId
where score > 70

30. 查询不及格的课程

SELECT 
    DISTINCT cid
FROM
    sc
WHERE
    score < 60

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

SELECT
    b.*
FROM
    (
        SELECT DISTINCT
            Sid
        FROM
            sc
        WHERE
            score >= 80
        AND Cid = '01'
    ) AS a
JOIN student AS b ON a.Sid = b.Sid

32. 求每门课程的学生人数

SELECT
    cid,
    COUNT(DISTINCT sid)
FROM
    sc
GROUP BY
    CId

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

SELECT
    *
FROM
    teacher AS a
JOIN course AS b ON a.TId = b.TId
AND a.Tname = '张三'
JOIN sc AS c ON b.CId = c.CId
JOIN student AS d ON d.SId = c.SId
HAVING
    max(c.score)

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

SELECT
    *
FROM
    teacher AS a
JOIN course AS b ON a.TId = b.TId
AND a.Tname = '张三'
JOIN sc AS c ON b.CId = c.CId
JOIN student AS d ON d.SId = c.SId
WHERE
    c.score = (
        SELECT
            max(c.score)
        FROM
            teacher AS a
        JOIN course AS b ON a.TId = b.TId
        AND a.Tname = '张三'
        JOIN sc AS c ON b.CId = c.CId
        JOIN student AS d ON d.SId = c.SId
    )

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

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

36. 查询每门功成绩最好的前两名

SELECT
    *
FROM
    sc
WHERE
    (
        SELECT
            count(*)
        FROM
            sc AS a
        WHERE
            sc.cid = a.cid
        AND sc.score < a.score
    ) < 3 ##是不是比自己大的有三条
ORDER BY
    cid ASC,
    sc.score DESC;

37. 统计每门课程的学生选修人数（超过 5 人的课程才统计）

SELECT
    CId,
    COUNT(DISTINCT SId) AS Qty
FROM
    sc
GROUP BY
    CId
HAVING
    Qty > 5

38. 检索至少选修两门课程的学生学号

SELECT
    SId,
    COUNT(DISTINCT CId) AS Qty
FROM
    sc
GROUP BY
    SId
HAVING
    Qty >= 2

39. 查询选修了全部课程的学生信息

SELECT
    b.*, COUNT(DISTINCT a.CId) AS Qty
FROM
    sc AS a
JOIN student AS b ON a.SId = b.SId
GROUP BY
    SId
HAVING
    Qty = (
        SELECT
            COUNT(DISTINCT CId)
        FROM
            course
    )

40. 查询各学生的年龄，只按年份来算

SELECT
    *, YEAR (NOW()) - YEAR (Sage) + 1 AS Age
FROM
    student

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

SELECT
    *, TIMESTAMPDIFF(YEAR, Sage, NOW()) AS Age
FROM
    student

42. 查询本周过生日的学生

SELECT
    *
FROM
    student
WHERE
    WEEKOFYEAR(Sage) = WEEKOFYEAR(NOW())

43. 查询下周过生日的学生

SELECT
    *
FROM
    student
WHERE
    WEEKOFYEAR(Sage) = WEEKOFYEAR(NOW())+1

44. 查询本月过生日的学生

SELECT
    *
FROM
    student
WHERE
    MONTH(Sage) = MONTH(NOW())

45. 查询下月过生日的学生

SELECT
    *
FROM
    student
WHERE
    MONTH(Sage) = MONTH(NOW())+1