最全50个Mysql数据库查询练习题

吾爱主题阅读：236 2024-04-02 07:58:42 评论：0

此数据库查询语句是网络上50个数据库查询练习题目，网上有些版本是oracle语句写的，大多数公司还是用免费的mysql数据库，以下都是mysql版本，全部都有验证过。

表名和字段

–1.学生表
Student(s#, sname, sage,ssex) –学生编号,学生姓名, 出生年月,学生性别
–2.课程表
Course(c#,cname,t#) – –课程编号, 课程名称, 教师编号
–3.教师表
Teacher(t#,tname) –教师编号,教师姓名
–4.成绩表
Sc(s#,c#,score) –学生编号,课程编号,分数

测试数据

用数据库可视化工具做练习非常方便，推荐使用sqlyog，软件图标是一只海豚。

在新连接种填上本机地址，用户名，密码和端口就直接连上mysql。

所有测试数据如下：

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 # --插入学生表测试数据 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' , '1990-05-20' , '男' ); INSERT INTO student VALUES ( '04' , '李云' , '1990-08-06' , '男' ); INSERT INTO student VALUES ( '05' , '周梅' , '1991-12-01' , '女' ); INSERT INTO student VALUES ( '06' , '吴兰' , '1992-03-01' , '女' ); INSERT INTO student VALUES ( '07' , '郑竹' , '1989-07-01' , '女' ); INSERT INTO student VALUES ( '08' , '王菊' , '1990-01-20' , '女' ); # --插入课程表测试数据 INSERT INTO course VALUES ( '01' , '语文' , '02' ); INSERT INTO course VALUES ( '02' , '数学' , '01' ); INSERT INTO course VALUES ( '03' , '英语' , '03' ); # --插入教师表测试数据 INSERT INTO teacher VALUES ( '01' , '张三' ); INSERT INTO teacher VALUES ( '02' , '李四' ); INSERT INTO teacher VALUES ( '03' , '王五' ); # --插入成绩表测试数据 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);

最后是50个数据库查询练习，已经验证过，是mysql版本的。

1. 查询" 01 "课程比" 02 "课程成绩高的学生的信息及课程分数

1 2 3 4 5 6 SELECT * FROM ( SELECT `s#` AS sno1, `c#` AS cno1, score FROM sc WHERE `c#`=01) a LEFT JOIN ( SELECT `s#` AS sno2, `c#` AS cno2, score FROM sc WHERE `c#`=02) b ON a.sno1 = b.sno2 WHERE a.score > b.score

1.1 查询同时存在" 01 "课程和" 02 "课程的情况

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 )

1 2 3 4 5 SELECT * FROM ( SELECT `s#` AS sno1, `c#` AS cno1, score FROM sc WHERE `c#`=01) a LEFT JOIN ( SELECT `s#` AS sno2, `c#` AS cno2, score FROM sc WHERE `c#`=02) b ON a.sno1 = b.sno2

1.3 查询不存在" 01 "课程但存在" 02 "课程的情况

SELECT * FROM
sc WHERE `c#`='02' AND `s#` NOT IN (SELECT `s#` FROM sc WHERE `c#`='01')

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

1 2 3 4 5 SELECT a.`s#`,b.`sname`, a.avg_score FROM ( SELECT `s#` , AVG (score) AS avg_score FROM sc GROUP BY `s#`) AS a LEFT JOIN student AS b ON a.`s#` = b.`s#` WHERE a.avg_score >=60

3. 查询在 SC 表存在成绩的学生信息

SELECT * FROM student WHERE `s#` IN (SELECT DISTINCT `s#` FROM sc)

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 )

1 2 3 4 5 SELECT `s#` ,sname , course_num , score_sum FROM ( SELECT `s#`, sname FROM student ) AS a LEFT JOIN ( SELECT `s#` AS sno , COUNT (`c#`) AS course_num , SUM (score) AS score_sum FROM sc GROUP BY sno) AS b ON a.`s#` = b.sno

4.1 查有成绩的学生信息

# 在最外面一层select的时候，不可以用函数
# 如果两张表连接之后，有相同的字段，这时候select就需要把其中一个字段改名

1 2 3 4 5 6 SELECT `s#` ,sname , course_num , score_sum FROM ( SELECT `s#`, sname FROM student ) AS a LEFT JOIN ( SELECT `s#` AS sno , COUNT (`c#`) AS course_num , SUM (score) AS score_sum FROM sc GROUP BY sno) AS b ON a.`s#` = b.sno WHERE course_num IS NOT NULL

5. 查询「李」姓老师的数量

SELECT COUNT(*) FROM teacher WHERE tname LIKE '李%'

6. 查询学过「张三」老师授课的同学的信息

1 2 3 4 5 # 张三老师是01号 SELECT * FROM student WHERE `s#` IN ( SELECT `s#` FROM sc WHERE `c#` = ( SELECT `c#` FROM course WHERE `t#` = ( SELECT `t#` FROM teacher WHERE tname= '张三' )))

# 7. 查询没有学全所有课程的同学的信息

SELECT `s#`,COUNT(`c#`) AS course_num FROM sc GROUP BY `s#`
HAVING course_num < (SELECT COUNT(*) 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

1 2 3 4 SELECT * FROM student WHERE `s#` IN ( SELECT DISTINCT `s#` FROM sc WHERE `c#` IN ( SELECT `c#` FROM sc WHERE `s#`=01)) AND `s#`!= 01

# 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

1 2 3 4 5 6 7 SELECT `s#` FROM ( SELECT * FROM sc LEFT JOIN ( SELECT `c#` AS cno FROM sc WHERE `s#` =01) a ON sc.`c#` = a.cno) AS b GROUP BY `s#` HAVING COUNT (b.`s#`) = ( SELECT COUNT (`c#`) AS cno FROM sc WHERE `s#` =01)

# 10. 查询没学过"张三"老师讲授的任一门课程的学生姓名

1 2 3 4 5 6 # 张三是01 # 01老师是教数学，c#是02 SELECT * FROM student WHERE `s#` NOT IN ( SELECT DISTINCT `s#` FROM sc WHERE `c#` IN ( SELECT `c#` FROM course WHERE `t#` IN ( SELECT `t#` FROM teacher WHERE tname = '张三' )))

# 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

1 2 3 4 5 6 7 8 SELECT `s#`, sname, avg_score FROM ( SELECT `s#`, sname FROM student WHERE `s#` IN ( SELECT a.`s#` FROM ( SELECT `s#`, COUNT (`c#`) AS num FROM sc WHERE score <60 GROUP BY `s#`) a WHERE num >=2)) AS b LEFT JOIN ( SELECT `s#` AS sno , AVG (score) AS avg_score FROM sc GROUP BY `s#`) AS c ON b.`s#` = c.sno

# 12. 检索" 01 "课程分数小于 60，按分数降序排列的学生信息

1 2 3 4 5 6 7 SELECT `s#`, sname, score FROM student AS a LEFT JOIN ( SELECT `s#` AS sno,`c#`,score FROM sc WHERE `c#`= 01 AND score <60 )b ON a.`s#`= b.sno WHERE score IS NOT NULL ORDER BY 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 `s#` ,AVG(score) AS avg_score FROM sc GROUP BY `s#` ORDER BY avg_score 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

1 2 3 4 5 6 7 8 9 10 11 SELECT DISTINCT a.`c#`,cname,最高分,最低分,平均分,及格率,中等率,优良率,优秀率 FROM sc a LEFT JOIN course ON a.`c#`=course.`c#` LEFT JOIN ( SELECT `c#`, MAX (score)最高分, MIN (score)最低分, AVG (score)平均分 FROM sc GROUP BY `c#`)b ON a.`c#`=b.`c#` LEFT JOIN ( SELECT `c#`, ROUND( r1 /cnt * 100, 2 ) AS 及格率 FROM ( SELECT `c#`, ( SUM ( CASE WHEN score >=60 THEN 1 ELSE 0 END )*1.00) AS r1 , COUNT (*) AS cnt FROM sc GROUP BY `c#`) c1) c ON a.`c#`=c.`c#` LEFT JOIN ( SELECT `c#`, ROUND( r2 /cnt * 100, 2 ) AS 中等率 FROM ( SELECT `c#`, ( SUM ( CASE WHEN score >=70 AND score<80 THEN 1 ELSE 0 END )*1.00) AS r2 , COUNT (*) AS cnt FROM sc GROUP BY `c#`) d1) d ON a.`c#`=d.`c#` LEFT JOIN ( SELECT `c#`, ROUND( r3 /cnt * 100, 2 ) AS 优良率 FROM ( SELECT `c#`, ( SUM ( CASE WHEN score >=80 AND score<90 THEN 1 ELSE 0 END )*1.00) AS r3 , COUNT (*) AS cnt FROM sc GROUP BY `c#`) e1) e ON a.`c#`=e.`c#` LEFT JOIN ( SELECT `c#`, ROUND( r4 /cnt * 100, 2 ) AS 优秀率 FROM ( SELECT `c#`, ( SUM ( CASE WHEN score >=90 THEN 1 ELSE 0 END )*1.00) AS r4 , COUNT (*) AS cnt FROM sc GROUP BY `c#`) f1) f ON a.`c#`=f.`c#`

# 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

1 2 3 4 5 6 7 8 9 10 11 # mysql中没有rank()函数 # 这种是重复时候保留名次，所以最后名次和人数是一样的 SELECT `s#`, `c#`, score, rank FROM ( SELECT `s#`, `c#`, score, @currank := IF(@prevrank = score, @currank, @incrank) AS rank, @incrank := @incrank + 1, @prevrank := score FROM sc , ( SELECT @currank :=0, @prevrank := NULL , @incrank := 1 ) r ORDER BY score DESC ) s

# 15.1 按各科成绩进行排序，并显示排名， Score 重复时合并名次

1 2 3 4 5 6 7 8 9 # 这种是当有重复名次的时候变成只有一个名次，所以排名的数量会变少 SELECT `s#`, `c#`, score, CASE WHEN @prevrank = score THEN @currank WHEN @prevrank := score THEN @currank := @currank + 1 END AS rank FROM sc, ( SELECT @currank :=0, @prevrank := NULL ) r ORDER BY score DESC

# 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

1 2 3 4 5 6 7 8 9 10 # from 后面不需要加表的别名 SELECT `s#`, sum_score, rank FROM ( SELECT `s#`, sum_score, @currank := IF(@prevrank = sum_score, @currank, @incrank) AS rank, @incrank := @incrank + 1, @prevrank := sum_score FROM ( SELECT `s#`, SUM (score) AS sum_score FROM sc GROUP BY `s#`) c , ( SELECT @currank :=0, @prevrank := NULL , @incrank := 1) r ORDER BY sum_score DESC ) s

# 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

1 2 3 4 5 6 7 8 9 10 11 SELECT c.*, CASE WHEN @prevrank = c.sum_score THEN @currank WHEN @prevrank := c.sum_score THEN @currank := @currank + 1 END AS rank FROM ( SELECT a.`s#`,a.sname, SUM (score) AS sum_score FROM (student AS a RIGHT JOIN sc AS b ON a.`s#` = b.`s#`) GROUP BY a.`s#` ) c , ( SELECT @currank := 0 , @prevrank := NULL ) d ORDER BY sum_score DESC

# 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

1 2 3 4 5 6 7 8 9 10 11 12 SELECT a.`c#` , b.cname, SUM ( CASE WHEN score >=85 AND score <=100 THEN 1 ELSE 0 END ) '[100-85]' , SUM ( CASE WHEN score >=85 AND score <=100 THEN 1 ELSE 0 END )*1.00/ COUNT (*) AS '[100-85]percent' , SUM ( CASE WHEN score < 85 AND score >= 70 THEN 1 ELSE 0 END ) '(85-70]' , SUM ( CASE WHEN score < 85 AND score >= 70 THEN 1 ELSE 0 END )*1.00/ COUNT (*) AS '(85-70]percent' , SUM ( CASE WHEN score < 70 AND score >= 60 THEN 1 ELSE 0 END ) '(70-60]' , SUM ( CASE WHEN score < 70 AND score >= 60 THEN 1 ELSE 0 END )*1.00/ COUNT (*) AS '(85-70]percent' , SUM ( CASE WHEN score < 60 AND score >= 0 THEN 1 ELSE 0 END ) '(60-0]' , SUM ( CASE WHEN score < 60 AND score >= 0 THEN 1 ELSE 0 END )*1.00/ COUNT (*) AS '(85-70]percent' , COUNT (*) AS counts FROM sc a LEFT JOIN course b ON a.`c#` = b.`c#` GROUP BY `c#`

# 18. 查询各科成绩前三名的记录

1 2 3 SELECT * FROM sc a WHERE ( SELECT COUNT (*) FROM sc WHERE `c#`=a.`c#` AND score>a.score)<3 ORDER BY a.`c#`, a.score DESC ;

# 19. 查询每门课程被选修的学生数

1 2 3 4 5 6 SELECT `c#`, COUNT (`s#`) FROM ( SELECT `s#`,`c#` FROM sc ORDER BY `c#`)a GROUP BY `c#` SELECT a.`c#` , b.cname , COUNT (*) AS num FROM sc a LEFT JOIN course b ON a.`c#` = b.`c#` GROUP BY a.`c#`;

# 20. 查询出只选修两门课程的学生学号和姓名

1 2 3 4 5 6 SELECT a.`s#`, a.sname ,cnt FROM student a LEFT JOIN ( SELECT `s#`, COUNT (`c#`) AS cnt FROM sc GROUP BY `s#`) b ON a.`s#`=b.`s#` WHERE cnt=2

# 21. 查询男生、女生人数

SELECT ssex,COUNT(ssex) 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

1 2 3 4 SELECT a.*,b.同名人数 FROM student a LEFT JOIN ( SELECT sname,ssex, COUNT (*) AS 同名人数 FROM student GROUP BY sname,ssex)b ON a.sname=b.sname AND a.ssex=b.ssex WHERE b.同名人数>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 `c#`, ROUND(AVG(score),2) AS avg_score FROM sc GROUP BY `c#` ORDER BY `c#` 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

1 2 3 4 5 6 7 8 9 SELECT c.`s#`,sname ,avg_score FROM (student c LEFT JOIN ( SELECT `s#`, avg_score FROM ( SELECT `s#` ,ROUND( AVG (score),2) AS avg_score FROM sc GROUP BY `s#` ORDER BY avg_score DESC )a WHERE avg_score >=85) b ON c.`s#` =b.`s#`) WHERE avg_score IS NOT NULL

# 27. 查询课程名称为「数学」，且分数低于 60 的学生姓名和分数

1 2 3 4 5 6 7 8 SELECT a.`s#`,a.sname,b.math, b.score FROM student a LEFT JOIN ( SELECT `s#`,`c#` AS math ,score FROM sc WHERE `c#` IN ( SELECT `c#` FROM course WHERE cname = '数学' ) AND sc.score <60) b ON a.`s#`=b.`s#` WHERE b.score IS NOT NULL

# 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）

1 2 3 SELECT a.`s#`,a.`sname`,a.`sage`,a.`ssex`,b.`c#`,b.score FROM student a LEFT JOIN sc b ON a.`s#` = b.`s#` LEFT JOIN course c ON c.`c#` = b.`c#`

# 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

1 2 3 4 5 6 7 SELECT a.`s#`,a.`sname`,a.`sage`,a.`ssex`,b.`c#`,b.score FROM student a LEFT JOIN ( SELECT `s#`,`c#`,score FROM sc WHERE score >70) b ON a.`s#`=b.`s#` LEFT JOIN course c ON c.`c#`=b.`c#` WHERE score IS NOT NULL

# 30. 查询不及格的课程

SELECT * 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

1 2 3 4 5 6 SELECT a.`s#`, a.sname ,b.score FROM student a LEFT JOIN ( SELECT * FROM sc WHERE `c#`= '01' AND score >= 80) b ON a.`s#` = b.`s#` WHERE score IS NOT NULL

# 32. 求每门课程的学生人数

SELECT `c#`,COUNT(`c#`) FROM sc GROUP BY `c#`

# 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

1 2 3 4 5 6 7 8 SELECT a.`s#`, a.`sname` ,b.`c#`, b.max_score FROM student a LEFT JOIN ( SELECT `s#` AS sid,`c#` , MAX (score) AS max_score FROM sc WHERE `c#` IN ( SELECT `c#` FROM course WHERE `t#` IN ( SELECT `t#` FROM teacher WHERE tname = '张三' ))) b ON a.`s#`=b.sid WHERE max_score IS NOT NULL

# 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

1 2 3 4 5 6 7 8 9 10 11 12 13 SELECT * FROM ( SELECT dd.*, CASE WHEN @prevrank = dd.score THEN @currank WHEN @prevrank := dd.score THEN @currank := @currank + 1 END AS rank FROM ( SELECT a.*,b.score FROM student a LEFT JOIN sc b ON a.`s#` = b.`s#` LEFT JOIN course c ON b.`c#` = c.`c#` LEFT JOIN teacher d ON c.`t#` = d.`t#` WHERE d.tname = '张三' ) dd,( SELECT @currank := 0 , @prevrank := NULL ) ff ORDER BY score DESC ) AS dddddddd WHERE rank = 1;

# 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

1 2 3 SELECT DISTINCT a.`s#`, a.`c#`, a.score FROM sc AS a JOIN sc AS b WHERE a.`c#` != b.`c#` AND a.score = b.score AND a.`s#` != b.`s#` ORDER BY a.`s#`, a.`c#`, a.score

# 36. 查询每门功课成绩最好的前两名

1 2 3 4 # 此题和18题相同 SELECT * FROM sc a WHERE ( SELECT COUNT (*) FROM sc WHERE `c#`=a.`c#` AND score>a.score)<2 ORDER BY a.`c#`, a.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）

1 2 3 4 5 # 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 a.`c#`, COUNT (*) AS num FROM course a LEFT JOIN sc b ON a.`c#` = b.`c#` GROUP BY a.`c#` HAVING num > 5 ORDER BY num,a.`c#`

# 38. 检索至少选修两门课程的学生学号

SELECT DISTINCT`s#`,COUNT(`c#`) AS num FROM sc GROUP BY `s#` HAVING num >=2

# 39. 查询选修了全部课程的学生信息

1 2 3 SELECT * FROM ( SELECT `s#`, COUNT (*) AS num FROM sc GROUP BY `s#` ) b WHERE num = ( SELECT COUNT (*) FROM course)

# 40. 查询各学生的年龄，只按年份来算

SELECT *, YEAR(NOW()) - YEAR(sage) AS age FROM student

# 41. 查询本周过生日的学生

1 2 3 4 SELECT * FROM ( SELECT * , WEEK(sage), MONTH (sage), DAY (sage), WEEK(STR_TO_DATE(CONCAT_WS( ',' , YEAR (NOW()), MONTH (sage), DAY (sage)), '%y,%m,%d' )) AS w FROM student) a WHERE w = WEEK(NOW())

# 42. 查询下周过生日的学生

1 2 3 4 SELECT * FROM ( SELECT * , WEEK(sage), MONTH (sage), DAY (sage),WEEK(NOW()), WEEK(STR_TO_DATE(CONCAT_WS( ',' , YEAR (NOW()), MONTH (sage), DAY (sage)), '%y,%m,%d' )) AS w FROM student) a WHERE w + 2 = WEEK(NOW())

# 43. 查询本月过生日的学生

1 2	`SELECT` `* ,` `MONTH` `(sage),` `MONTH` `(NOW())` `FROM` `student` `WHERE` `MONTH` `(sage) =` `MONTH` `(NOW())`

# 44. 查询下月过生日的学生

1 2	`SELECT` `* ,` `MONTH` `(sage),` `MONTH` `(NOW())` `FROM` `student` `WHERE` `MONTH` `(sage) =` `MONTH` `(NOW()) + 1`

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