一、准备数据
数据表结构
1.学生表：
Student(s_id,s_name,s_birth,s_sex) --学生编号,学生姓名, 出生年月,学生性别
2.课程表：
Course(c_id,c_name,t_id) – --课程编号, 课程名称, 教师编号
3.教师表：
Teacher(t_id,t_name) --教师编号,教师姓名
4.成绩表：
Score(s_id,c_id,s_score) --学生编号,课程编号,分数
创建4张表
1、创建学生表
CREATE TABLE Student(
s_id VARCHAR(20) NOT NULL PRIMARY KEY COMMENT '学生编号',
s_name VARCHAR(40) COMMENT '学生姓名',
s_birth VARCHAR(40) COMMENT '出生年月',
s_sex VARCHAR(4) COMMENT '学生性别'
);
2、创建课程表
CREATE TABLE Course(
c_id VARCHAR(20) NOT NULL PRIMARY KEY COMMENT '课程编号',
c_name VARCHAR(40) COMMENT '课程名称',
t_id VARCHAR(20) COMMENT '教师编号'
);
3、创建教师表teacher
CREATE TABLE teacher(
t_id VARCHAR(20) NOT NULL PRIMARY KEY COMMENT '教师编号',
t_name VARCHAR(40) COMMENT '教师姓名'
);
4、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(10) COMMENT '分数',
PRIMARY KEY(s_id,c_id)
);
插入测试数据
为了便于大家交流，这里借用网上经典数据 1、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');
2、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');
3、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');
4、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);
二、mysql经典50题
1、查询"01"课程比"02"课程成绩高的学生的信息及课程分数
SELECT a.*,b.`s_score` AS 01_score,c.`s_score` AS 02_score FROM student 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' WHERE b.`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
SELECT a.*,b.`s_score` AS 01_score,c.`s_score` AS 02_score FROM student 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' WHERE b.`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
SELECT a.s_id AS id ,a.s_name AS `name`,ROUND(AVG(b.s_score),2) AS avg_score FROM student a JOIN score b
ON a.`s_id`=b.`s_id` GROUP BY b.s_id HAVING avg_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)
SELECT a.s_id AS id,a.s_name AS `name`, ROUND(AVG(b.s_score),2) AS avg_score FROM student a LEFT JOIN score b ON a.`s_id`=b.`s_id` GROUP BY b.`s_id` HAVING avg_score<60
UNION
SELECT s_id AS id,s_name AS `name`, 0 AS avg_score FROM student
WHERE 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
select s.s_id,s.s_name,count(s1.s_score) '选课总数',sum(IFNULL(s1.s_score,0)) total_score
from student s
left join score s1 on s1.s_id=s.s_id
GROUP BY s.s_id
order by total_score DESC;
6、查询"李"姓老师的数量
SELECT COUNT(*) FROM teacher WHERE t_name LIKE '李%';
7、查询学过"张三"老师授课的同学的信息
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 IN (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
SELECT * FROM Student WHERE s_id NOT IN
(SELECT s_id FROM score WHERE c_id IN
(SELECT c_id FROM course WHERE t_id IN (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
SELECT a.*,b.s_score AS 01_score,c.s_score AS 02_score FROM student 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';
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 * FROM student
WHERE s_id IN (SELECT DISTINCT s_id FROM score WHERE c_id='01')
AND s_id NOT IN (SELECT DISTINCT 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 * FROM student
WHERE s_id NOT IN(
SELECT s_id FROM score
GROUP BY s_id HAVING COUNT(*) =(SELECT COUNT(DISTINCT c_id) FROM course));
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 * FROM student WHERE s_id IN
(SELECT DISTINCT s_id FROM score WHERE c_id IN (SELECT c_id FROM score WHERE 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
方法1
SELECT * FROM student WHERE s_id IN
(SELECT s_id FROM score WHERE c_id IN (SELECT c_id FROM score WHERE s_id='01')
GROUP BY s_id HAVING COUNT(1)=(SELECT COUNT(1)FROM score WHERE s_id='01'))
AND s_id!='01';
方法2
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_id
HAVING COUNT(1)=(SELECT COUNT(1) FROM score WHERE s_id='01'));
方法3 自己写的
SELECT * from student stu
where stu.s_id in (
select temp.s_id from
(
select sc1.s_id, GROUP_CONCAT(sc1.c_id) AS res from score sc1 GROUP BY sc1.s_id
HAVING res=(select GROUP_CONCAT(sc.c_id) from score sc GROUP BY sc.s_id HAVING sc.s_id='01')
AND sc1.s_id<>'01'
) temp
)
14、查询没学过"张三"老师讲授的任一门课程的学生姓名
select * from student stu where stu.s_id not in(
select sc.s_id from score sc WHERE sc.c_id in
(select c.c_id from teacher t,course c where t.t_id=c.t_id and t_name='张三')
);
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 第一个是本人方法，感觉可读性更好
select stu.s_name,stu.s_id,temp.avg_score from student stu , (
select distinct sc.s_id,round(AVG(sc.s_score),2) avg_score from score sc where sc.s_score<60 GROUP BY sc.s_id) as temp where stu.s_id=temp.s_id
网上的方法
SELECT a.s_id,a.s_name,ROUND(AVG(b.s_score),2) AS avg_score FROM student a JOIN
(SELECT * FROM score WHERE s_score<60)b ON a.s_id=b.s_id GROUP BY a.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
SELECT a.*,b.s_score FROM Student a JOIN Score b ON a.`s_id`=b.`s_id`
WHERE b.`c_id`='01' AND b.`s_score`<60 ORDER BY b.`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1
select
stu.s_id, stu.s_name,
IFNULL(sc1.s_score,0) '数学',
IFNULL(sc2.s_score,0) '语文',
IFNULL(sc3.s_score,0) '英语',
ROUND((IFNULL(sc1.s_score,0)+IFNULL(sc2.s_score,0)+IFNULL(sc3.s_score,0))/3,2) '平均成绩'
from student stu
LEFT JOIN score sc1 on sc1.c_id='01' and sc1.s_id=stu.s_id
LEFT JOIN score sc2 on sc2.c_id='02' and sc2.s_id=stu.s_id
LEFT JOIN score sc3 on sc3.c_id='03' and sc3.s_id=stu.s_id
order by 平均成绩 desc
方法1查询结果：
方法2
SELECT
a.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2查询结果：
方法3【方法3最好，要掌握】
SELECT a.s_id,MAX(CASE a.c_id WHEN '01' THEN a.s_score END ) 语文,
MAX(CASE a.c_id WHEN '02' THEN a.s_score END ) 数学,
MAX(CASE a.c_id WHEN '03' THEN a.s_score END ) 英语,
AVG(a.s_score),b.s_name FROM Score a JOIN Student b ON a.s_id=b.s_id GROUP BY a.s_id ORDER BY 5 DESC;
方法3查询结果：
方法1是我自己写的，方法2和方法3是网上的，我感觉我写更合理，自行比较吧！
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
SELECT a.`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
FROM score a JOIN course b
ON a.`c_id`=b.`c_id`
GROUP BY a.`c_id`;
19、按各科成绩进行排序，并显示排名
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 score
FROM (
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)s
UNION
SELECT 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 score
FROM (
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)s
UNION
SELECT 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 score
FROM (
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)a,
(SELECT @k:=0,@i:=0,@score:=0)s;
21、查询不同老师所教不同课程平均分从高到低显示
SELECT c.`t_name`,a.`c_id`,ROUND(AVG(a.`s_score`),2) AS avg_score
FROM score a JOIN course b ON a.`c_id`=b.`c_id` JOIN teacher c ON b.`t_id`=c.`t_id`
GROUP BY c.`t_name`,a.`c_id`
ORDER BY avg_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
m.*,n.c_id,n.rank,n.score
FROM student m,
(SELECT a.s_id,a.c_id,
@i:=@i+1 AS i,
@k:=(CASE WHEN @score=a.s_score THEN @K ELSE @k+1 END ) AS rank,
@score:=a.s_score AS score
FROM
(SELECT * FROM score WHERE c_id='01' ORDER BY c_id,s_score DESC) a,(SELECT @i:=0,@k:=0,@score:=0) s) n
WHERE m.`s_id`=n.s_id AND n.rank IN (2,3)
UNION
SELECT
m.*,n.c_id,n.rank,n.score
FROM student m,
(SELECT a.s_id,a.c_id,
@i1:=@i1+1 AS i,
@k1:=(CASE WHEN @score1=a.s_score THEN @K1 ELSE @k1+1 END ) AS rank,
@score1:=a.s_score AS score
FROM
(SELECT * FROM score WHERE c_id='02' ORDER BY c_id,s_score DESC) a,(SELECT @i1:=0,@k1:=0,@score1:=0) s) n
WHERE m.`s_id`=n.s_id AND n.rank IN (2,3)
UNION
SELECT
m.*,n.c_id,n.rank,n.score
FROM student m,
(SELECT a.s_id,a.c_id,
@i2:=@i2+1 AS i,
@k2:=(CASE WHEN @score2=a.s_score THEN @K2 ELSE @k2+1 END ) AS rank,
@score2:=a.s_score AS score
FROM
(SELECT * FROM score WHERE c_id='03' ORDER BY c_id,s_score DESC) a,(SELECT @i2:=0,@k2:=0,@score2:=0) s) n
WHERE m.`s_id`=n.s_id AND n.rank IN (2,3);
之所以两个union的三部分中，@i、@i1、@i2要做区分，因为每部分都要取rank=2和3，这与19题是不同的
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
SELECT a.`c_id`, b.`c_name` ,
SUM(CASE WHEN a.s_score>=85 THEN 1 ELSE 0 END) AS '[100-85]',
ROUND(100*(SUM(CASE WHEN a.s_score>=85 THEN 1 ELSE 0 END)/COUNT(1)),2) AS '[100-85]占比',
SUM(CASE WHEN a.s_score<85 AND a.s_score>=70 THEN 1 ELSE 0 END) AS '[85-70]',
ROUND(100*(SUM(CASE WHEN a.s_score<85 AND a.s_score>=70 THEN 1 ELSE 0 END)/COUNT(1)),2) AS '[85-70]占比',
SUM(CASE WHEN a.`s_score`<70 AND a.`s_score`>=60 THEN 1 ELSE 0 END) AS '[70-60]',
ROUND(100*(SUM(CASE WHEN a.`s_score`<70 AND a.`s_score`>=60 THEN 1 ELSE 0 END)/COUNT(1)),2) AS '[70-60]占比',
SUM(CASE WHEN a.`s_score`<60 THEN 1 ELSE 0 END) AS '[0-60]',
ROUND(100*(SUM(CASE WHEN a.`s_score`<60 THEN 1 ELSE 0 END)/COUNT(1)),2) AS '[0-60]占比'
FROM score a JOIN course b ON a.c_id=b.c_id GROUP BY a.c_id;
24、查询学生平均成绩及其名次
SELECT a.s_id,
@i:=@i+1 AS 'i不保留空缺排名',
@k:=(CASE WHEN @score=a.avg_score THEN @k ELSE @i END) AS 'rank保留空缺排名',
@score:=a.avg_score AS score
FROM (SELECT s_id,ROUND(AVG(s_score),2) AS avg_score FROM score GROUP BY s_id ORDER BY avg_score DESC) a,
(SELECT @i:=0,@k:=0,@score:=0) s;
25、查询各科成绩前三名的记录
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 JOIN score b ON a.`s_id`=b.`s_id`
GROUP BY b.`s_id` HAVING COUNT(1)=2;
方法2
SELECT s_id,s_name FROM student WHERE s_id IN
(SELECT s_id FROM score GROUP BY s_id HAVING COUNT(c_id)=2);
28、查询男生、女生人数
SELECT s_sex AS 性别,COUNT(1) AS 人数 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 s_birth LIKE '1990%';
方法2
SELECT s_name FROM student WHERE YEAR(s_birth)='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) AS avg_score FROM score GROUP BY c_id ORDER BY avg_score DESC,c_id;
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 avg_score
FROM student a JOIN score b ON a.`s_id`=b.`s_id` GROUP BY b.`s_id` HAVING avg_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 JOIN score b
ON a.`s_id`=b.`s_id` AND c_id=(SELECT c_id FROM course WHERE c_name='数学') WHERE b.`s_score`<60;
35、查询所有学生的课程及分数情况；
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;
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 思路：先把score表中分数>70的部分筛选出来，然后再与student、course两张表连表查询
SELECT a.s_name,c.`c_name`,b.s_score FROM student a
JOIN (SELECT s_id,c_id,s_score FROM score WHERE s_score>70) b ON a.`s_id`=b.s_id
JOIN course c ON b.c_id=c.`c_id`;
另外一种思路
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 JOIN score b ON a.`s_id`=b.`s_id`
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 a.s_id,a.s_name,b.`s_score` FROM student a JOIN score b
ON a.`s_id`=b.`s_id` WHERE b.`c_id`='01' AND b.`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 c.*,d.s_score FROM student c JOIN score d ON c.`s_id`=d.`s_id` AND d.`s_score` IN
(SELECT MAX(s_score) FROM score WHERE c_id IN
(SELECT a.c_id FROM course a JOIN teacher b ON a.`t_id`=b.`t_id` AND b.`t_name`='张三'));
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 JOIN score b
ON a.`s_score`=b.`s_score` AND a.`c_id`!=b.`c_id`;
注意DISTINCT关键字不能丢
42、查询每门功课成绩最好的前两名
方法1
(SELECT s_id,c_id,s_score FROM score WHERE c_id='01' ORDER BY s_score DESC LIMIT 2 )
UNION
(SELECT s_id,c_id,s_score FROM score WHERE c_id='02' ORDER BY s_score DESC LIMIT 2 )
UNION
(SELECT s_id,c_id,s_score FROM score WHERE c_id='03' ORDER BY s_score DESC LIMIT 2 );
方法2（此方法是网上的代码，笔者觉得不错，也列在这里供大家交流参考。）
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 选修人数 DESC ,课程号;
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
方法1
SELECT a.* FROM student a JOIN score b ON a.`s_id`=b.`s_id`
GROUP BY b.`s_id` HAVING COUNT(1)=(SELECT COUNT(c_id) FROM course);
方法2
SELECT * FROM student WHERE s_id IN(
SELECT s_id FROM score GROUP BY s_id HAVING COUNT(*)=(SELECT COUNT(*) 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
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
SELECT * FROM student WHERE WEEK(NOW())=WEEK(s_birth);
48、查询下周过生日的学生
SELECT * FROM student WHERE WEEK(s_birth)=WEEK(NOW())+1;
49、查询本月过生日的学生
方法1
SELECT * FROM student WHERE MONTH(NOW())=MONTH(s_birth);
方法2
SELECT * FROM student WHERE DATE_FORMAT(NOW(),'%m')=DATE_FORMAT(s_birth,'%m');
50、查询下月过生日的学生
SELECT * FROM student WHERE MONTH(NOW())+1=MONTH(s_birth);