MySQL经典练习50题
2024-06-04 15:52:40
这50道查询练习确实很经典，题是我在网上找的，SQL是我自己写的。发现问题的欢迎提出，有更好方法的，可以提出来大家共同学习。做完后你的SQL编写能力肯定有一个提升。
---------创建数据库、表、插入数据------------------------ 建表
-- 学生表
CREATE TABLE Student(s_id VARCHAR(20) COMMENT '学生编号',s_name VARCHAR(20) NOT NULL DEFAULT '' COMMENT '学生姓名',s_birth VARCHAR(20) NOT NULL DEFAULT '' COMMENT '出生年月',s_sex VARCHAR(10) NOT NULL DEFAULT '' COMMENT '学生性别',PRIMARY KEY(s_id)
)ENGINE=InnoDB DEFAULT CHARSET=utf8 COMMENT '学生表';
-- 课程表
CREATE TABLE Course(c_id VARCHAR(20) COMMENT '课程编号',c_name VARCHAR(20) NOT NULL DEFAULT '' COMMENT '课程名称',t_id VARCHAR(20) NOT NULL COMMENT '教师编号',PRIMARY KEY(c_id)
)ENGINE=InnoDB DEFAULT CHARSET=utf8 COMMENT '课程表';
-- 教师表
CREATE TABLE Teacher(t_id VARCHAR(20) COMMENT '教师编号',t_name VARCHAR(20) NOT NULL DEFAULT '' COMMENT '教师姓名',PRIMARY KEY(t_id)
)ENGINE=InnoDB DEFAULT CHARSET=utf8 COMMENT '教师表';
-- 成绩表
CREATE TABLE Score(s_id VARCHAR(20) COMMENT '学生编号',c_id VARCHAR(20) COMMENT '课程编号',s_score INT(3) COMMENT '分数',PRIMARY KEY(s_id,c_id)
)ENGINE=InnoDB DEFAULT CHARSET=utf8 COMMENT '成绩表';-- 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');
-- 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');-- 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');-- 成绩表测试数据
insert into Score values('01' , '01' , 80);
insert into Score values('01' , '02' , 90);
insert into Score values('01' , '03' , 99);
insert into Score values('02' , '01' , 70);
insert into Score values('02' , '02' , 60);
insert into Score values('02' , '03' , 80);
insert into Score values('03' , '01' , 80);
insert into Score values('03' , '02' , 80);
insert into Score values('03' , '03' , 80);
insert into Score values('04' , '01' , 50);
insert into Score values('04' , '02' , 30);
insert into Score values('04' , '03' , 20);
insert into Score values('05' , '01' , 76);
insert into Score values('05' , '02' , 87);
insert into Score values('06' , '01' , 31);
insert into Score values('06' , '03' , 34);
insert into Score values('07' , '02' , 89);
insert into Score values('07' , '03' , 98);---------------------------------------------------- 1、查询"01"课程比"02"课程成绩高的学生的信息及课程分数
SELECTa.*, b.s_score AS score1,c.s_score AS score2
FROMstudent a
LEFT JOIN score b ON a.s_id = b.s_id
AND b.c_id = '01'
LEFT JOIN score c ON a.s_id = c.s_id
AND (c.c_id = '02' OR c.c_id =NULL)
WHEREb.s_score > c.s_score ;-- 2、查询"01"课程比"02"课程成绩低的学生的信息及课程分数
SELECTa.*, b.s_score AS score1,c.s_score AS score2
FROMstudent a
LEFT JOIN score b ON a.s_id = b.s_id
AND (b.c_id = '01' OR b.c_id =NULL)
LEFT JOIN score c ON a.s_id = c.s_id
AND c.c_id = '02'
WHEREb.s_score < c.s_score ;-- 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
SELECTa.s_id,a.s_name,ROUND(AVG(b.s_score), 1) AS 平均成绩
FROMstudent a
LEFT JOIN score b ON a.s_id = b.s_id
GROUP BYa.s_id
HAVINGAVG(b.s_score) >= 60;-- 4、查询平均成绩小于60分的同学的学生编号和学生姓名和平均成绩
-- (包括有成绩的和无成绩的)
-- 方法一：
SELECTa.s_id,a.s_name,ROUND(AVG(b.s_score), 1) AS 平均成绩
FROMstudent a
LEFT JOIN score b ON a.s_id = b.s_id
GROUP BYa.s_id
HAVINGAVG(b.s_score) < 60
OR
a.s_id NOT IN(SELECT DISTINCT a.s_id FROM student a JOIN score b WHERE a.s_id=b.s_id);-- 方法二：
SELECTa.s_id,a.s_name,ROUND(AVG(b.s_score),2) AS avg_score
FROMstudent a
JOIN score b ON a.s_id = b.s_id
GROUP BYa.s_id
HAVING AVG(b.s_score) < 60
UNION
SELECT a.s_id,a.s_name,0 AS avg_score FROM
student a
WHERE a.s_id NOT IN (SELECT DISTINCT s_id FROM score);-- 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,并从高到低排序
SELECTa.s_id,a.s_name,COUNT(b.c_id) AS 选课总数 ,SUM(b.s_score) AS 总成绩
FROMstudent a
LEFT JOIN score b
ONa.s_id = b.s_id
GROUP BY a.s_id
ORDER BY SUM(b.s_score) DESC;-- 6、查询"李"姓老师的数量
SELECT COUNT(t_id) FROM teacher WHERE t_name LIKE '李%';-- 7、查询学过"张三"老师授课的同学的信息
#张三编号
SELECT t_id FROM teacher WHERE t_name = '张三'
#张三代课的课程编号
SELECT c_id FROM course WHERE t_id = (SELECT t_id FROM teacher WHERE t_name = '张三')
#学张三课程的学生编号
SELECT s_id FROM score WHERE c_id = (SELECT c_id FROM course WHERE t_id = (SELECT t_id FROM teacher WHERE t_name = '张三'))
-- 方法一：
SELECT *FROM student WHERE
s_id IN (SELECT s_id FROM score WHERE
c_id = (SELECT c_id FROM course WHERE
t_id = (SELECT t_id FROM teacher WHERE t_name = '张三'))
);-- 方法二：
SELECT a.* FROM student a
JOIN score b ON a.s_id = b.s_id
WHERE
b.c_id IN (SELECT c_id FROM course WHERE t_id = (SELECT t_id FROM teacher WHERE t_name = '张三'));-- 8、查询没学过"张三"老师授课的同学的信息
-- 方法一：
SELECT *FROM student WHERE
s_id NOT IN (SELECT s_id FROM score WHERE
c_id = (SELECT c_id FROM course WHERE
t_id = (SELECT t_id FROM teacher WHERE t_name = '张三'))
);-- 9、查询学过编号为"01"并且也学过编号为"02"的课程的同学的信息,及两门课程成绩
-- 方法一
SELECTa.*, b.s_score,c.s_score
FROMstudent a
JOIN score b ON a.s_id = b.s_id
AND b.c_id = '01'
JOIN score c ON a.s_id = c.s_id
AND c.c_id = '02';-- 方法二：
SELECTa.*, b.s_score,c.s_score
FROMstudent a,score b,score c
WHEREa.s_id = b.s_id
AND a.s_id = c.s_id
AND b.c_id = '01'
AND c.c_id = '02';-- 10、查询学过编号为"01"但是没有学过编号为"02"的课程的同学的信息
SELECT s_id FROM score WHERE c_id = '01'
SELECT s_id FROM score WHERE c_id = '02'SELECT *FROM student a
WHERE
a.s_id IN (SELECT s_id FROM score WHERE c_id = '01')
AND a.s_id NOT IN (SELECT s_id FROM score WHERE c_id = '02');-- 11、查询没有学全所有课程的同学的信息 SELECT s_id FROM score
GROUP BY s_id
HAVING COUNT(s_id) != 3#方法一：
SELECT *FROM student WHERE s_id IN(SELECT s_id FROM scoreGROUP BY s_idHAVING COUNT(s_id) != 3
);#方法二：
SELECT a.s_id FROM score a JOIN score b ON a.s_id=b.s_id AND b.c_id='02'JOIN score c ON a.s_id=c.s_id AND c.c_id='03'WHERE a.c_id='01'SELECT *FROM student d WHERE d.s_id IN(SELECT e.s_id FROM score e WHERE e.s_id NOT IN(SELECT a.s_id FROM score a JOIN score b ON a.s_id=b.s_id AND b.c_id='02'JOIN score c ON a.s_id=c.s_id AND c.c_id='03'WHERE a.c_id='01')
);#--------------# 上述两种方法结果都少了没有选课的8号学生，但看具体条件是否需要查出# 学全选取所有课程的同学的id
SELECT s_id FROM scoreGROUP BY s_idHAVING COUNT(s_id) = 3#方法一：
SELECT *FROM student WHERE s_id NOT IN(SELECT s_id FROM scoreGROUP BY s_idHAVING COUNT(s_id) = 3
);-- 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 c_id FROM score WHERE s_id ='01'SELECT DISTINCT s_id FROM score WHERE c_id IN (SELECT c_id FROM score WHERE s_id ='01')SELECT * FROM student a
WHERE
a.s_id IN (SELECT DISTINCT b.s_id FROM score b WHERE
b.c_id IN (SELECT c.c_id FROM score c WHERE c.s_id ='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 * FROM student WHERE s_id IN(SELECT DISTINCT s_id FROM score WHERE s_id!='01' AND c_id IN (SELECT c_id FROM score WHERE s_id ='01') GROUP BY s_idHAVING COUNT(1)=(SELECT COUNT(1) FROM score WHERE s_id='01')
)-- 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 s_id FROM score WHERE s_score<60 GROUP BY s_id HAVING COUNT(1) >= 2SELECTa.s_id,a.s_name,ROUND(AVG(b.s_score), 1) AS 平均成绩
FROMstudent a
LEFT JOIN score b ON a.s_id = b.s_id
GROUP BY a.s_id
HAVING
a.s_id IN(SELECT s_id FROM score WHERE s_score<60 GROUP BY s_id HAVING COUNT(1) >= 2)-- 16、检索"01"课程分数小于60，按分数降序排列的学生信息及01分数
-- 方法一：
SELECTa.*, b.s_score
FROMstudent a
LEFT JOIN score b ON a.s_id = b.s_id
WHEREb.c_id = '01'
AND b.s_score < 60
ORDER BYb.s_score DESC;-- 方法二：
SELECTa.*, b.s_score
FROMstudent a,score b
WHERE
a.s_id = b.s_id AND b.c_id='01' AND b.s_score < 60
ORDER BYb.s_score DESC;-- 方法三（有点瑕疵）：
SELECTa.*, b.s_score
FROMstudent a
LEFT JOIN score b ON a.s_id = b.s_id
AND b.c_id = '01'
AND b.s_score < 60
ORDER BYb.s_score DESC;-- 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-- 方法一（自连接）：
SELECTa.s_id,
(SELECT s_score FROM score WHERE s_id=a.s_id AND c_id='01') AS score1,
(SELECT s_score FROM score WHERE s_id=a.s_id AND c_id='02') AS score2,
(SELECT s_score FROM score WHERE s_id=a.s_id AND c_id='03') AS score3,
ROUND(avg(a.s_score), 2) AS 平均分
FROM score a
GROUP BY a.s_id
ORDER BY 平均分 DESC;-- 方法二（自连接）：
SELECT a.s_id,b.s_score,c.s_score,d.s_score,ROUND(avg(a.s_score), 2) AS 平均分
FROMscore a
LEFT JOIN score b ON a.s_id = b.s_id AND b.c_id='01'
LEFT JOIN score c ON a.s_id = c.s_id AND c.c_id='02'
LEFT JOIN score d ON a.s_id = d.s_id AND d.c_id='03'
GROUP BY a.s_id
ORDER BY 平均分 DESC;-- 18.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：>=90SELECT
a.c_id,
MAX(a.s_score),MIN(a.s_score),AVG(a.s_score)
FROMscore a
GROUP BY a.c_id;-- 方法一：
SELECTa.c_id AS 课程ID,b.c_name AS 课程name,MAX(a.s_score) AS 最高分,MIN(a.s_score) AS 最低分,ROUND(AVG(a.s_score),2) AS 平均分,ROUND(100*(SUM(CASE WHEN a.s_score >= 60 THEN 1 ELSE 0 END) / COUNT(1)) , 2) AS '及格率',ROUND(100*(SUM(CASE WHEN a.s_score >= 70 AND a.s_score <80 THEN 1 ELSE 0 END) / COUNT(1)) , 2) AS '中等率',ROUND(100*(SUM(CASE WHEN a.s_score >= 80 AND a.s_score <90 THEN 1 ELSE 0 END) / COUNT(1)) , 2) AS '优良率',ROUND(100*(SUM(CASE WHEN a.s_score >= 90 THEN 1 ELSE 0 END) / COUNT(1)) , 2) AS '优秀率'
FROMscore a
LEFT JOIN course b ON a.c_id = b.c_id
GROUP BYb.c_id;-- 19、按各科成绩进行排序，并显示排名
-- mysql没有rank顺序函数select a.s_id,a.c_id,@i:=@i +1 as i保留排名,@k:=(case when @score=a.s_score then @k else @i end) as rank不保留排名,@score:=a.s_score as scorefrom (select s_id,c_id,s_score from score WHERE c_id='01' GROUP BY s_id,c_id,s_score ORDER BY s_score DESC
)a,(select @k:=0,@i:=0,@score:=0)sunionselect a.s_id,a.c_id,@i:=@i +1 as i,@k:=(case when @score=a.s_score then @k else @i end) as rank,@score:=a.s_score as scorefrom (select s_id,c_id,s_score from score WHERE c_id='02' GROUP BY s_id,c_id,s_score ORDER BY s_score DESC
)a,(select @k:=0,@i:=0,@score:=0)sunionselect a.s_id,a.c_id,@i:=@i +1 as i,@k:=(case when @score=a.s_score then @k else @i end) as rank,@score:=a.s_score as scorefrom (select s_id,c_id,s_score from score WHERE c_id='03' GROUP BY s_id,c_id,s_score ORDER BY s_score DESC
)a,(select @k:=0,@i:=0,@score:=0)s-- 20、查询学生的总成绩并进行排名
SELECT a.s_id,@i:=@i+1 AS i,@k:=(CASE WHEN @score=a.sum_score THEN @k ELSE @i END) AS rank,@score:=a.sum_score AS score
FROM (SELECT s_id,SUM(s_score) AS sum_score FROM score GROUP BY s_id ORDER BY sum_score DESC) AS a,
(SELECT @i:=0,@score:=0) AS b-- 21、查询不同老师所教不同课程平均分从高到低显示
SELECTa.t_name,b.c_id,b.c_name,ROUND(AVG(c.s_score) ,2) AS 平均分
FROMteacher a
LEFT JOIN course b ON a.t_id = b.t_id
LEFT JOIN score c ON b.c_id=c.c_id
GROUP BY c.c_id
ORDER BY AVG(c.s_score) DESC;-- 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
SELECT c.*,d.s_name,d.s_birth,d.s_sex FROM
(SELECT a.s_id,a.s_score,a.c_id,@i:=@i+1 AS 排名 FROM score a,(SELECT @i:=0)b WHERE a.c_id='01') c
LEFT JOIN student d ON c.s_id=d.s_id
WHERE 排名 BETWEEN 2 AND 3
UNION
SELECT c.*,d.s_name,d.s_birth,d.s_sex FROM
(SELECT a.s_id,a.s_score,a.c_id,@j:=@j+1 AS 排名 FROM score a,(SELECT @j:=0)b WHERE a.c_id='02') c
LEFT JOIN student d ON c.s_id=d.s_id
WHERE 排名 BETWEEN 2 AND 3
UNION
SELECT c.*,d.s_name,d.s_birth,d.s_sex FROM
(SELECT a.s_id,a.s_score,a.c_id,@k:=@k+1 AS 排名 FROM score a,(SELECT @k:=0)b WHERE a.c_id='03') c
LEFT JOIN student d ON c.s_id=d.s_id
WHERE 排名 BETWEEN 2 AND 3;-- 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SELECTa.c_id AS 课程编号, a.c_name AS 课程名称,c.`[100-85]的人数`, c.`[100-85]所占百分比`,d.`[85-70]的人数`, d.`[85-70]所占百分比`,e.`[70-60]的人数`, e.`[70-60]所占百分比`,f.`[0-60]的人数`, f.`[0-60]所占百分比`
FROMcourse a
LEFT JOIN score b ON a.c_id = b.c_id
LEFT JOIN(SELECT *,SUM(CASE WHEN s_score >85 AND s_score <=100 THEN 1 ELSE 0 END) AS '[100-85]的人数' ,ROUND(SUM(CASE WHEN s_score >85 AND s_score <=100 THEN 1 ELSE 0 END)/COUNT(1)*100,2) AS '[100-85]所占百分比'FROM score GROUP BY c_id) c ON a.c_id=c.c_id
LEFT JOIN(SELECT*,SUM(CASE WHEN s_score >70 AND s_score <=85 THEN 1 ELSE 0 END) AS '[85-70]的人数' ,ROUND(SUM(CASE WHEN s_score >70 AND s_score <=85 THEN 1 ELSE 0 END)/COUNT(1)*100,2) AS '[85-70]所占百分比'FROM score GROUP BY c_id) d ON a.c_id=d.c_id
LEFT JOIN(SELECT*,SUM(CASE WHEN s_score >60 AND s_score <=70 THEN 1 ELSE 0 END) AS '[70-60]的人数' ,ROUND(SUM(CASE WHEN s_score >60 AND s_score <=70 THEN 1 ELSE 0 END)/COUNT(1)*100,2) AS '[70-60]所占百分比'FROM score GROUP BY c_id) e ON a.c_id=e.c_id
LEFT JOIN(SELECT *,SUM(CASE WHEN s_score >0 AND s_score <=60 THEN 1 ELSE 0 END) AS '[0-60]的人数' ,ROUND(SUM(CASE WHEN s_score >0 AND s_score <=60 THEN 1 ELSE 0 END)/COUNT(1)*100,2) AS '[0-60]所占百分比'FROM score GROUP BY c_id) f ON a.c_id=f.c_id
GROUP BY a.c_id-- 24、查询学生平均成绩及其名次
SELECTb.s_id,@i:=@i+1 AS 相同分数的不同名次,@k:=(CASE WHEN @avg_s=b.avg_score THEN @k ELSE @i END) AS 相同分数的相同名次,@avg_s:=b.avg_score AS 平均成绩
FROM
(SELECTa.s_id,ROUND(AVG(a.s_score), 2) AS avg_score
FROMscore a
GROUP BYa.s_id
ORDER BY AVG(a.s_score) DESC) b,(SELECT @i:=0,@avg_s:=0,@k:=0) c-- 24.1添加名次rank,（相同分数的相同名次，并列排名）
-- 上面24难以看出并列排名
SELECTb.s_id, b.c_id,-- 顺序一直在变大@i:=@i+1 AS 相同分数的不同名次,-- 只有在前后二次分数不同时才会使用顺序号@k:=(CASE WHEN @s=b.s_score THEN @k ELSE @i END) AS 相同分数的相同名次,@s:=b.s_score AS 成绩
FROM
(SELECT *FROM score WHERE s_id='03' ORDER BY s_score DESC)b,
(SELECT @i:=0,@k:=0,@s:=0)c-- 25、查询各科成绩前三名的记录
-- 1.选出b表比a表成绩大的所有组
-- 2.选出比当前id成绩大的 小于三个的
-- SELECT a.s_id,a.c_id,a.s_score FROM score a
-- LEFT JOIN score b ON a.c_id=b.c_id AND a.s_score<b.s_score
-- GROUP BY a.s_id,a.c_id,a.s_score
-- HAVING COUNT(b.s_id)<3
-- ORDER BY a.c_id,a.s_score DESC-- 26、查询每门课程被选修的学生数
SELECT c_id,COUNT(1) FROM score GROUP BY c_id-- 27、查询出只有两门课程的全部学生的学号和姓名
-- 方法一：
SELECT a.s_id,a.s_name FROM student a
LEFT JOIN score b ON a.s_id=b.s_id
GROUP BY a.s_id
HAVING COUNT(1)=2-- 方法二：
SELECT a.s_id,a.s_name FROM student a WHERE a.s_id IN
(SELECT s_id FROM score GROUP BY s_id HAVING COUNT(1)=2)-- 28、查询男生、女生人数
SELECT s_sex,COUNT(1) FROM student GROUP BY s_sex-- 29、查询名字中含有"风"字的学生信息
SELECT *FROM student WHERE s_name LIKE '%风%' -- 30、查询同名同性学生名单，并统计同名人数
SELECT a.s_name,a.s_sex,COUNT(1) AS 人数 FROM student a
JOIN student b ON a.s_name=b.s_name AND a.s_sex=b.s_sex AND a.s_id!=b.s_id
GROUP BY a.s_name,a.s_sex-- 31、查询1990年出生的学生名单
-- 方法一
SELECT s_name FROM student WHERE YEAR(s_birth)='1990'
-- 方法二
SELECT s_name FROM student WHERE s_birth LIKE '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 c_id,ROUND(avg(s_score),2)FROM score
GROUP BY c_id
ORDER BY avg(s_score) DESC,c_id ASC -- 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.s_id,a.s_name,ROUND(avg(b.s_score),2) AS 平均成绩 FROM student a
LEFT JOIN score b ON a.s_id=b.s_id
GROUP BY a.s_id
HAVING avg(b.s_score) >= 85-- 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 a.s_name,b.s_score FROM student a
LEFT JOIN score b ON a.s_id=b.s_id
WHERE c_id=(
SELECT c_id FROM course WHERE c_name='数学'
) AND b.s_score < 60-- 35、查询所有学生的课程及分数情况；
-- 方法一：
SELECT a.s_id,a.s_name,b.s_score,c.s_score,d.s_score FROM student a
LEFT JOIN score b ON a.s_id=b.s_id AND b.c_id='01'
LEFT JOIN score c ON a.s_id=c.s_id AND c.c_id='02'
LEFT JOIN score d ON a.s_id=d.s_id AND d.c_id='03'
GROUP BY a.s_id-- 方法二：SELECT a.s_id,a.s_name,
SUM(CASE c.c_name WHEN '语文' THEN b.s_score ELSE 0 END) AS '语文',
SUM(CASE c.c_name WHEN '数学' THEN b.s_score ELSE 0 END) AS '数学',
SUM(CASE c.c_name WHEN '英语' THEN b.s_score ELSE 0 END) AS '英语',
SUM(b.s_score) as '总分'
FROM student a
LEFT JOIN score b ON a.s_id = b.s_id
LEFT JOIN course c ON b.c_id = c.c_id
GROUP BY a.s_id,a.s_name-- 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 a.s_name,c.c_name,b.s_score FROM student a
LEFT JOIN score b ON a.s_id=b.s_id
LEFT JOIN course c ON b.c_id=c.c_id
HAVING b.s_score > 70-- 37、查询不及格的学生id,姓名，及其课程名称，分数
SELECT a.s_id,a.s_name,c.c_name,b.s_score FROM student a
LEFT JOIN score b ON a.s_id=b.s_id
LEFT JOIN course c ON b.c_id=c.c_id
WHERE b.s_score < 60-- 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 b.s_id,b.s_name FROM score a
LEFT JOIN student b ON a.s_id=b.s_id
WHERE a.c_id='01' AND a.s_score>80-- 39、求每门课程的学生人数
SELECT c_id,COUNT(1) FROM score GROUP BY c_id-- 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SELECT t_id FROM teacher a WHERE a.t_name='张三'SELECT c_id FROM course b WHERE t_id=(SELECT t_id FROM teacher a WHERE a.t_name='张三')SELECT c.*,d.s_score FROM student c
LEFT JOIN score d ON c.s_id=d.s_id AND d.c_id=
(SELECT c_id FROM course b WHERE t_id=(SELECT t_id FROM teacher a WHERE a.t_name='张三'))
HAVING MAX(d.s_score)-- 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
SELECT DISTINCT a.s_id,a.c_id,a.s_score
FROM score a,score b WHERE a.s_score=b.s_score AND a.c_id!=b.c_id-- 42、查询每门功课成绩最好的前两名
-- 方法一
SELECT *FROM
(SELECT a.s_id,a.c_id,a.s_score, @i:=@i+1 as 排名 FROM score a,(SELECT @i:=0)b WHERE a.c_id='01' ORDER BY a.s_score DESC
) c
WHERE 排名 BETWEEN 1 AND 2
UNION
SELECT *FROM
(SELECT a.s_id,a.c_id,a.s_score, @j:=@j+1 as 排名 FROM score a,(SELECT @j:=0)b WHERE a.c_id='02' ORDER BY a.s_score DESC
) c
WHERE 排名 BETWEEN 1 AND 2
UNION
SELECT *FROM
(SELECT a.s_id,a.c_id,a.s_score, @k:=@k+1 as 排名 FROM score a,(SELECT @k:=0)b WHERE a.c_id='03' ORDER BY a.s_score DESC
) c
WHERE 排名 BETWEEN 1 AND 2-- 方法二
-- SELECT a.s_id,a.c_id,a.s_score FROM score a
-- WHERE (SELECT COUNT(1) FROM score b WHERE b.c_id=a.c_id AND b.s_score>=a.s_score)<=2 ORDER BY a.c_id-- 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 c_id AS 课程号,COUNT(1) AS 选修人数 FROM score
GROUP BY c_id
HAVING COUNT(1)>5
ORDER BY COUNT(1) DESC,c_id-- 44、检索至少选修两门课程的学生学号
SELECT s_id,COUNT(1) FROM score GROUP BY s_id HAVING COUNT(1)>=2-- 45、查询选修了全部课程的学生信息
SELECT COUNT(1) FROM courseSELECT b.* FROM score a
LEFT JOIN student b ON a.s_id=b.s_id
GROUP BY a.s_id
HAVING COUNT(1)=(SELECT COUNT(1) FROM course)-- 46、查询各学生的年龄-- 按照出生日期来算，当前月日<出生年月的月日则，年龄减一
-- 方法一
SELECT s_id,s_birth,(DATE_FORMAT(NOW(),'%Y')-DATE_FORMAT(s_birth,'%Y'))-
(CASE WHEN DATE_FORMAT(NOW(),'%m%d')< DATE_FORMAT(s_birth,'%m%d') THEN 1 ELSE 0 END) AS age
FROM student-- 47、查询本周过生日的学生
-- 方法一
SELECT * FROM student WHERE WEEK(CURRENT_DATE)=WEEK(s_birth)
-- 方法二
SELECT * FROM student WHERE WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))=WEEK(s_birth)-- 48、查询下周过生日的学生
SELECT * FROM student WHERE WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))+1=WEEK(s_birth)-- 49、查询本月过生日的学生
SELECT * FROM student WHERE MONTH(NOW())=MONTH(s_birth)-- 50、查询下月过生日的学生
SELECT * FROM student WHERE MONTH(NOW())+1=MONTH(s_birth)SELECT DATE_FORMAT(NOW(),'%Y')
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