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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 6SELECT * 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 3 4 5 6SELECT * 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 sno2 IS NOT NULL
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 5SELECT * 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 5SELECT 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 5SELECT `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 6SELECT `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 4SELECT * 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 7SELECT `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 8SELECT `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 7SELECT `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 11SELECT 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 11SELECT 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 12SELECT 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 3SELECT * 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 6SELECT `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 6SELECT 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 4SELECT 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 9SELECT 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 8SELECT 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 3SELECT 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 7SELECT 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 6SELECT 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 8SELECT 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 13SELECT * 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 3SELECT 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 3SELECT * 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 4SELECT * 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 4SELECT * 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 2SELECT * , MONTH(sage),MONTH(NOW()) FROM student
WHERE MONTH(sage) = MONTH(NOW())
# 44. 查询下月过生日的学生
? 1 2SELECT * , MONTH(sage),MONTH(NOW()) FROM student
WHERE MONTH(sage) = MONTH(NOW()) + 1
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原文链接：https://blog.csdn.net/youmianzhou/article/details/86562649
