MYSQL-习题掌握

文章目录

  • SQL基本操作
  • 1 设计表操作
    • 1.1 关系表字段
    • 1.2 关系表创建
    • 1.3 关系表数据
    • 1.4 关系表关系
  • 2 SQL操作
    • 2.1 SQL 1-10
    • 2.2 SQL 11-20
    • 2.3 SQL 21-30
    • 2.4 SQL 31-40
    • 2.5 SQL 41-50

SQL基本操作

1 设计表操作

1.1 关系表字段

  • 1 学生表 student
s_id s_name s_birth s_sex
学生编号 学生姓名 学生年月 学生性别
  • 2 课程表 course
c_id c_name t_id
课程编号 课程名称 教师标号
  • 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 课程分数

1.2 关系表创建

  • 1 创建 student 表
CREATE TABLE student(
s_id VARCHAR(20),
s_name VARCHAR(20) NOT NULL DEFAULT '',
s_birth VARCHAR(20) NOT NULL DEFAULT '',
s_sex VARCHAR(10) NOT NULL DEFAULT '',
PRIMARY KEY(s_id)
) ENGINE=INNODB DEFAULT CHARSET=utf8;
  • 2 创建 course 表
CREATE TABLE course(
c_id VARCHAR(20),
c_name VARCHAR(20) NOT NULL DEFAULT '',
t_id VARCHAR(20) NOT NULL,
PRIMARY KEY(c_id)
) ENGINE=INNODB DEFAULT CHARSET=utf8;
  • 3 创建 teacher 表
CREATE TABLE teacher(
t_id VARCHAR(20),
t_name VARCHAR(20) NOT NULL DEFAULT '',
PRIMARY KEY(t_id)
) ENGINE=INNODB DEFAULT CHARSET=utf8;
  • 4 创建score表
CREATE TABLE score(
s_id VARCHAR(20),
c_id VARCHAR(20),
s_score INT(3),
PRIMARY KEY(s_id,c_id)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

1.3 关系表数据

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');
INSERT INTO course VALUES('01' , '语文' , '02');
INSERT INTO course VALUES('02' , '数学' , '01');
INSERT INTO course VALUES('03' , '英语' , '03');
INSERT INTO teacher VALUES('01' , '张雪峰');
INSERT INTO teacher VALUES('02' , '李佳琦');
INSERT INTO teacher VALUES('03' , '王思聪');
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);

1.4 关系表关系

MYSQL-习题掌握_第1张图片

2 SQL操作

2.1 SQL 1-10

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'
LEFT JOIN Score c 
ON a.s_id=c.s_id AND c.c_id='02' OR c.c_id = NULL 
WHERE b.s_score>c.s_score;

在这里插入图片描述

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 
LEFT JOIN score b 
ON a.s_id=b.s_id AND b.c_id='01' OR b.c_id=NULL
JOIN score c 
ON a.s_id=c.s_id AND c.c_id='02' 
WHERE b.s_score<c.s_score;

在这里插入图片描述

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 b.s_id,b.s_name,ROUND(AVG(a.s_score),2) AS avg_score 
FROM student b
JOIN score a ON b.s_id = a.s_id
GROUP BY b.s_id,b.s_name 
HAVING ROUND(AVG(a.s_score),2)>=60;

MYSQL-习题掌握_第2张图片

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 b.s_id,b.s_name,ROUND(AVG(a.s_score),2) AS avg_score 
FROM student b
LEFT JOIN score a 
ON b.s_id = a.s_id
GROUP BY b.s_id,b.s_name 
HAVING ROUND(AVG(a.s_score),2)<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);

在这里插入图片描述

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

SELECT a.s_id,a.s_name,
COUNT(b.c_id) AS sum_course,
SUM(b.s_score) AS sum_score 
FROM student a
LEFT JOIN score b 
ON a.s_id=b.s_id
GROUP BY a.s_id,a.s_name;

MYSQL-习题掌握_第3张图片

6、查询"李"姓老师的数量

SELECT COUNT(t_id) FROM teacher WHERE t_name LIKE '李%';

在这里插入图片描述

7、查询学过"张雪峰"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 =(
SELECT t_id FROM teacher WHERE t_name = '张雪峰'));

MYSQL-习题掌握_第4张图片

8、查询没学过"张雪峰"老师授课的同学的信息

SELECT c.* 
FROM student c
WHERE c.s_id NOT IN(
SELECT a.s_id 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 = '张雪峰')));

在这里插入图片描述

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

SELECT a.* 
FROM student a,score b,score c
WHERE a.s_id = b.s_id 
AND a.s_id = c.s_id 
AND b.c_id='01' 
AND c.c_id='02';

MYSQL-习题掌握_第5张图片

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 a.* 
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')

在这里插入图片描述

2.2 SQL 11-20

11、查询没有学全所有课程的同学的信息

SELECT s.* 
FROM student s 
WHERE s.s_id IN(
SELECT s_id FROM score WHERE 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'))

在这里插入图片描述

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 a.s_id FROM score a WHERE a.c_id IN(
SELECT a.c_id FROM score a WHERE a.s_id='01'));

MYSQL-习题掌握_第6张图片

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
方式1:

SELECT * FROM student WHERE s_id IN(
SELECT DISTINCT d.s_id FROM (
SELECT b.s_id , c.c_id FROM (
SELECT s.s_id,COUNT(a.c_id) c_num 
FROM student s,score a 
WHERE s.s_id = a.s_id 
GROUP BY s.s_id 
HAVING c_num=(SELECT COUNT(c_id) FROM score WHERE s_id='01')) b 
JOIN score c 
WHERE b.s_id = c.s_id AND c.c_id IN (
SELECT c_id FROM score WHERE s_id='01')) d) 
AND s_id !='01'; 
第一步:先查询s_id是01的学生的课程号和课程号的个数
SELECT COUNT(c_id) FROM score WHERE s_id='01'
SELECT c_id FROM score WHERE s_id='01'
第二步:求学生c_id的个数是3的s_id,并且c_id在第一步查询的c_id的值内
SELECT b.s_id , c.c_id FROM (
SELECT s.s_id,COUNT(a.c_id) c_num 
FROM student s,score a 
WHERE s.s_id = a.s_id 
GROUP BY s.s_id 
HAVING c_num=(
SELECT COUNT(c_id) 
FROM score WHERE s_id='01')) b 
JOIN score c WHERE b.s_id = c.s_id AND c.c_id IN (
SELECT c_id FROM score WHERE s_id='01')
第三步:查询第二步的结果中的s_id
第四步:查询学生信息且s_id在第三步的s_id中 并且去除掉编号是01的学生信息

方式2:

SELECT * 
FROM student 
WHERE s_id IN ( 
SELECT s_id FROM (
SELECT s_id, COUNT( s_id ) cou 
FROM ( 
SELECT * FROM score WHERE s_id IN (
SELECT s_id FROM (
SELECT s_id, COUNT( s_id ) COUNT 
FROM score WHERE s_id != '01' GROUP BY s_id ) t1 
WHERE t1.count = (
SELECT COUNT( c_id ) FROM score WHERE s_id = '01' ))) t3 
WHERE t3.c_id IN (
SELECT c_id FROM score WHERE s_id = '01' ) GROUP BY s_id ) t4 
WHERE t4.cou=(SELECT COUNT( c_id ) FROM score WHERE s_id = '01'))

在这里插入图片描述

– 14、查询没学过"张三"老师讲授的任一门课程的学生姓名

select a.s_name from student a where a.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’))
group by s_id);

– 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 a.s_id,a.s_name,ROUND(AVG(b.s_score)) from
student a
left join score b on a.s_id = b.s_id
where a.s_id in(
select s_id from score where s_score<60 GROUP BY s_id having count(1)>=2)
GROUP BY a.s_id,a.s_name

– 16、检索"01"课程分数小于60,按分数降序排列的学生信息

select a.*,b.c_id,b.s_score from
student a,score b
where a.s_id = b.s_id and 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

select a.s_id,(select s_score from score where s_id=a.s_id and
c_id=‘01’) as 语文,
(select s_score from score where s_id=a.s_id and c_id=‘02’) as 数学,
(select s_score from score where s_id=a.s_id and c_id=‘03’) as 英语,
round(avg(s_score),2) as 平均分 from score a 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:>=90

select
a.c_id,b.c_name,MAX(s_score),MIN(s_score),ROUND(AVG(s_score),2),
ROUND(100*(SUM(case when a.s_score>=60 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 及格率,
ROUND(100*(SUM(case when a.s_score>=70 and a.s_score<=80 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 中等率,
ROUND(100*(SUM(case when a.s_score>=80 and a.s_score<=90 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 优良率,
ROUND(100*(SUM(case when a.s_score>=90 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 优秀率
from score a left join course b on a.c_id = b.c_id GROUP BY a.c_id,b.c_name

– 19、按各科成绩进行排序,并显示排名(实现不完全)
– mysql没有rank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 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

2.3 SQL 21-30

– 21、查询不同老师所教不同课程平均分从高到低显示

select a.t_id,c.t_name,a.c_id,ROUND(avg(s_score),2) as avg_score from course a
left join score b on a.c_id=b.c_id
left join teacher c on a.t_id=c.t_id
GROUP BY a.c_id,a.t_id,c.t_name ORDER BY avg_score DESC;
1
2
3
4
– 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 d.,c.排名,c.s_score,c.c_id from (
select a.s_id,a.s_score,a.c_id,@i:=@i+1 as 排名 from score a,(select @i:=0)s 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 d.,c.排名,c.s_score,c.c_id from (
select a.s_id,a.s_score,a.c_id,@j:=@j+1 as 排名 from score a,(select @j:=0)s 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 d.*,c.排名,c.s_score,c.c_id from (
select a.s_id,a.s_score,a.c_id,@k:=@k+1 as 排名 from score a,(select @k:=0)s 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

select distinct
f.c_name,a.c_id,b.85-100,b.百分比,c.70-85,c.百分比,d.60-70,d.百分比,e.0-60,e.百分比
from score a
left join (select c_id,SUM(case when s_score >85 and s_score <=100 then 1 else 0 end) as 85-100,
ROUND(100*(SUM(case when s_score >85 and s_score <=100 then 1 else 0 end)/count()),2) as 百分比
from score GROUP BY c_id)b on a.c_id=b.c_id
left join (select c_id,SUM(case when s_score >70 and s_score <=85 then 1 else 0 end) as 70-85,
ROUND(100(SUM(case when s_score >70 and s_score <=85 then 1 else 0 end)/count()),2) as 百分比
from score GROUP BY c_id)c on a.c_id=c.c_id
left join (select c_id,SUM(case when s_score >60 and s_score <=70 then 1 else 0 end) as 60-70,
ROUND(100(SUM(case when s_score >60 and s_score <=70 then 1 else 0 end)/count()),2) as 百分比
from score GROUP BY c_id)d on a.c_id=d.c_id
left join (select c_id,SUM(case when s_score >=0 and s_score <=60 then 1 else 0 end) as 0-60,
ROUND(100(SUM(case when s_score >=0 and s_score <=60 then 1 else 0 end)/count(*)),2) as 百分比
from score GROUP BY c_id)e on a.c_id=e.c_id
left join course f on a.c_id = f.c_id

– 24、查询学生平均成绩及其名次

select a.s_id,
@i:=@i+1 as ‘不保留空缺排名’,
@k:=(case when @avg_score=a.avg_s then @k else @i end) as ‘保留空缺排名’,
@avg_score:=avg_s as ‘平均分’
from (select s_id,ROUND(AVG(s_score),2) as avg_s from score GROUP BY s_id)a,(select @avg_score:=0,@i:=0,@k:=0)b;

– 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

1
2
3
4
– 26、查询每门课程被选修的学生数

select c_id,count(s_id) from score a GROUP BY c_id

1
– 27、查询出只有两门课程的全部学生的学号和姓名

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);
1
2
– 28、查询男生、女生人数

select s_sex,COUNT(s_sex) as 人数  from student GROUP BY s_sex

1
– 29、查询名字中含有"风"字的学生信息

select * from student where s_name like '%风%';

1
– 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(*) from student a  JOIN 
            student b on a.s_id !=b.s_id and a.s_name = b.s_name and a.s_sex = b.s_sex
GROUP BY a.s_name,a.s_sex

1
2
3

2.4 SQL 31-40

– 31、查询1990年出生的学生名单

select s_name from student where s_birth like ‘1990%’
1
– 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 ASC
1
– 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,b.s_name,ROUND(avg(a.s_score),2) as avg_score from score a
left join student b on a.s_id=b.s_id GROUP BY s_id HAVING avg_score>=85
1
2
– 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 score b LEFT JOIN student a on a.s_id=b.s_id where b.c_id=(
            select c_id from course where c_name ='数学') and b.s_score<60

1
2
– 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,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,b.c_name,c.s_score from course b left join score c
on b.c_id = c.c_id
left join student a on a.s_id=c.s_id where c.s_score>=70

– 37、查询不及格的课程

select a.s_id,a.c_id,b.c_name,a.s_score from score a left join course
b on a.c_id = b.c_id
where a.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,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
1
2
– 39、求每门课程的学生人数

select count(*) 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

– 查询老师id
select c_id from course c,teacher d where c.t_id=d.t_id and d.t_name=‘张三’
– 查询最高分(可能有相同分数)
select MAX(s_score) from score where c_id=‘02’
– 查询信息
select a.*,b.s_score,b.c_id,c.c_name 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.c_id =(select c_id from course c,Teacher d where c.t_id=d.t_id and d.t_name=‘张三’)
and b.s_score in (select MAX(s_score) from Score where c_id=(select c_id from course c,Teacher d where c.t_id=d.t_id and
d.t_name=‘张三’))

2.5 SQL 41-50

– 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 b.s_id,b.c_id,b.s_score from score a,score b where a.c_id != b.c_id and a.s_score = b.s_score
1
– 42、查询每门功成绩最好的前两名
– 牛逼的写法

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,count(*) as total from score GROUP BY c_id HAVING total>5
ORDER BY total,c_id ASC

– 44、检索至少选修两门课程的学生学号

select s_id,count(*) as sel from score GROUP BY s_id HAVING sel>=2

– 45、查询选修了全部课程的学生信息

select * from student where s_id in(        
    select s_id from score GROUP BY s_id HAVING count(*)=(select count(*) from course))

1
2
–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_birth,(DATE_FORMAT(NOW(),’%Y’)-DATE_FORMAT(s_birth,’%Y’) -
(case when DATE_FORMAT(NOW(),’%m%d’)>DATE_FORMAT(s_birth,’%m%d’) then 0 else 1
end)) as age
from student;

– 47、查询本周过生日的学生

select * from student where WEEK(DATE_FORMAT(NOW(),‘%Y%m%d’))=WEEK(s_birth)
select * from student where YEARWEEK(s_birth)=YEARWEEK(DATE_FORMAT(NOW(),‘%Y%m%d’))
select WEEK(DATE_FORMAT(NOW(),‘%Y%m%d’))
1
2
3
– 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)
1
– 49、查询本月过生日的学生

select * from student where MONTH(DATE_FORMAT(NOW(),’%Y%m%d’))
=MONTH(s_birth)

– 50、查询下月过生日的学生

select * from student where MONTH(DATE_FORMAT(NOW(),‘%Y%m%d’))+1 =MONTH(s_birth)

你可能感兴趣的:(MySQL,mysql)