Mysql Sql 语句练习题（50道）

表名和字段

–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

测试数据

--建表
--学生表
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`)
);
--课程表
CREATE TABLE `Course`(
`c_id` VARCHAR(20),
`c_name` VARCHAR(20) NOT NULL DEFAULT '',
`t_id` VARCHAR(20) NOT NULL,
PRIMARY KEY(`c_id`)
);
--教师表
CREATE TABLE `Teacher`(
`t_id` VARCHAR(20),
`t_name` VARCHAR(20) NOT NULL DEFAULT '',
PRIMARY KEY(`t_id`)
);
--成绩表
CREATE TABLE `Score`(
`s_id` VARCHAR(20),
`c_id` VARCHAR(20),
`s_score` INT(3),
PRIMARY KEY(`s_id`,`c_id`)
);
--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);

表数据如下

student 学生表：

s_id s_name s_birth s_sex 01 赵雷 1990-01-01 男 02 钱电 1990-12-21 男 03 孙凤 1990-05-20 男 04 李云 1990-08-06 男 05 周梅 1991-12-12 女 06 吴兰 2017-12-13 女 07 郑竹 1989-07-01 女 08 王菊 1990-01-20 女 09 赵雷 1990-01-21 女 10 赵雷 1990-01-22 男

score 分数表：

s_id c_id s_score 01 01 80 01 02 90 01 03 99 02 01 70 02 02 60 02 03 80 03 01 80 03 02 80 03 03 80 04 01 50 04 02 30 04 03 20 05 01 76 05 03 87 06 01 31 06 03 34 07 03 89 07 01 98

course 课程表

c_id c_name t_id 01 语文 02 02 数学 01 03 英语 03

teacher 老师表：

t_id t_name 01 张三 02 李四 03 王五

-- 准备条件,去掉 sql_mode 的 ONLY_FULL_GROUP_BY 否则此种情况下会报错:
-- Expression #1 of select list is not in group by clause and contains nonaggregated column 'userinfo.
-- 原因：
-- MySQL 5.7.5和up实现了对功能依赖的检测。如果启用了only_full_group_by SQL模式(在默认情况下是这样)，
-- 那么MySQL就会拒绝选择列表、条件或顺序列表引用的查询，这些查询将引用组中未命名的非聚合列，而不是在功能上依赖于它们。
-- (在5.7.5之前，MySQL没有检测到功能依赖项，only_full_group_by在默认情况下是不启用的。关于前5.7.5行为的描述，请参阅MySQL 5.6参考手册。)
-- 执行以下个命令，可以查看 sql_mode 的内容。
SHOW SESSION VARIABLES;
SHOW GLOBAL VARIABLES;
select @@sql_mode;
-- 更改
set global sql_mode='STRICT_TRANS_TABLES,NO_ZERO_IN_DATE,NO_ZERO_DATE,ERROR_FOR_DIVISION_BY_ZERO,NO_AUTO_CREATE_USER,NO_ENGINE_SUBSTITUTION';
set session sql_mode='STRICT_TRANS_TABLES,NO_ZERO_IN_DATE,NO_ZERO_DATE,ERROR_FOR_DIVISION_BY_ZERO,NO_AUTO_CREATE_USER,NO_ENGINE_SUBSTITUTION';

练习题和sql

-- 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 st.*,sc.s_score as '语文' ,sc2.s_score '数学' 
from student st
left join score sc on sc.s_id=st.s_id and sc.c_id='01' 
left join score sc2 on sc2.s_id=st.s_id and sc2.c_id='02' 
where sc.s_score>sc2.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 st.*,sc.s_score '语文',sc2.s_score '数学' from student st
left join score sc on sc.s_id=st.s_id and sc.c_id='01'
left join score sc2 on sc2.s_id=st.s_id and sc2.c_id='02'
where sc.s_score<sc2.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 st.s_id,st.s_name,ROUND(AVG(sc.s_score),2) cjScore from student st
left join score sc on sc.s_id=st.s_id
group by st.s_id having AVG(sc.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)
select st.s_id,st.s_name,(case when ROUND(AVG(sc.s_score),2) is null then 0 else ROUND(AVG(sc.s_score)) end ) cjScore from student st
left join score sc on sc.s_id=st.s_id
group by st.s_id having AVG(sc.s_score)<60 or AVG(sc.s_score) is NULL

-- 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 st.s_id,st.s_name,count(c.c_id),( case when SUM(sc.s_score) is null or sum(sc.s_score)="" then 0 else SUM(sc.s_score) end) from student st
left join score sc on sc.s_id =st.s_id 
left join course c on c.c_id=sc.c_id
group by st.s_id

-- 6、查询"李"姓老师的数量 
select t.t_name,count(t.t_id) from teacher t
group by t.t_id having t.t_name like "李%"; 

-- 7、查询学过"张三"老师授课的同学的信息 
select st.* from student st 
left join score sc on sc.s_id=st.s_id
left join course c on c.c_id=sc.c_id
left join teacher t on t.t_id=c.t_id
 where t.t_name="张三"

-- 8、查询没学过"张三"老师授课的同学的信息 
 -- 张三老师教的课
 select c.* from course c left join teacher t on t.t_id=c.t_id where t.t_name="张三"
 -- 有张三老师课成绩的st.s_id
 select sc.s_id from score sc where sc.c_id in (select c.c_id from course c left join teacher t on t.t_id=c.t_id where t.t_name="张三")
 -- 不在上面查到的st.s_id的学生信息,即没学过张三老师授课的同学信息
 select st.* from student st where st.s_id not in(
 select sc.s_id from score sc where sc.c_id in (select c.c_id from course c left join teacher t on t.t_id=c.t_id where t.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
select st.* from student st 
inner join score sc on sc.s_id = st.s_id
inner join course c on c.c_id=sc.c_id and c.c_id="01"
where st.s_id in (
select st2.s_id from student st2 
inner join score sc2 on sc2.s_id = st2.s_id
inner join course c2 on c2.c_id=sc2.c_id and c2.c_id="02"
)


网友提供的思路(厉害呦~):
SELECT st.*
FROM student st
INNER JOIN score sc ON sc.`s_id`=st.`s_id`
GROUP BY st.`s_id`
HAVING SUM(IF(sc.`c_id`="01" OR sc.`c_id`="02" ,1,0))>1

-- 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 st.* from student st 
inner join score sc on sc.s_id = st.s_id
inner join course c on c.c_id=sc.c_id and c.c_id="01"
where st.s_id not in (
select st2.s_id from student st2 
inner join score sc2 on sc2.s_id = st2.s_id
inner join course c2 on c2.c_id=sc2.c_id and c2.c_id="02"
)

-- 11、查询没有学全所有课程的同学的信息
 -- 太复杂,下次换一种思路,看有没有简单点方法
 -- 此处思路为查学全所有课程的学生id,再内联取反面
select * from student where s_id not in (
select st.s_id from student st 
inner join score sc on sc.s_id = st.s_id and sc.c_id="01"
where st.s_id in (
select st2.s_id from student st2 
inner join score sc2 on sc2.s_id = st2.s_id and sc2.c_id="02"
) and st.s_id in (
select st2.s_id from student st2 
inner join score sc2 on sc2.s_id = st2.s_id and sc2.c_id="03"
))
-- 来自一楼网友的思路，左连接，根据学生id分组过滤掉 数量小于 课程表中总课程数量的结果(show me his code),简洁不少。
select st.* from Student st
left join Score S
on st.s_id = S.s_id
group by st.s_id
having count(c_id)<(select count(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 distinct st.* from student st 
left join score sc on sc.s_id=st.s_id
where sc.c_id in (
select sc2.c_id from student st2
left join score sc2 on sc2.s_id=st2.s_id
where st2.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 st.* from student st 
left join score sc on sc.s_id=st.s_id
group by st.s_id
having group_concat(sc.c_id) = 
(
select group_concat(sc2.c_id) from student st2
left join score sc2 on sc2.s_id=st2.s_id
where st2.s_id ='01'
)

-- 14、查询没学过"张三"老师讲授的任一门课程的学生姓名
select st.s_name from student st 
where st.s_id not in (
select sc.s_id from score sc 
inner join course c on c.c_id=sc.c_id
inner join teacher t on t.t_id=c.t_id and t.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
select st.s_id,st.s_name,avg(sc.s_score) from student st
left join score sc on sc.s_id=st.s_id
where sc.s_id in (
select sc.s_id from score sc 
where sc.s_score<60 or sc.s_score is NULL
group by sc.s_id having COUNT(sc.s_id)>=2
)
group by st.s_id

-- 16、检索"01"课程分数小于60，按分数降序排列的学生信息
select st.*,sc.s_score from student st 
inner join score sc on sc.s_id=st.s_id and sc.c_id="01" and sc.s_score<60
order by sc.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
 -- 可加round,case when then else end 使显示更完美
select st.s_id,st.s_name,avg(sc4.s_score) "平均分",sc.s_score "语文",sc2.s_score "数学",sc3.s_score "英语" from student st
left join score sc on sc.s_id=st.s_id and sc.c_id="01"
left join score sc2 on sc2.s_id=st.s_id and sc2.c_id="02"
left join score sc3 on sc3.s_id=st.s_id and sc3.c_id="03"
left join score sc4 on sc4.s_id=st.s_id
group by st.s_id 
order by SUM(sc4.s_score) 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 c.c_id,c.c_name,max(sc.s_score) "最高分",MIN(sc2.s_score) "最低分",avg(sc3.s_score) "平均分" 
,((select count(s_id) from score where s_score>=60 and c_id=c.c_id )/(select count(s_id) from score where c_id=c.c_id)) "及格率"
,((select count(s_id) from score where s_score>=70 and s_score<80 and c_id=c.c_id )/(select count(s_id) from score where c_id=c.c_id)) "中等率"
,((select count(s_id) from score where s_score>=80 and s_score<90 and c_id=c.c_id )/(select count(s_id) from score where c_id=c.c_id)) "优良率"
,((select count(s_id) from score where s_score>=90 and c_id=c.c_id )/(select count(s_id) from score where c_id=c.c_id)) "优秀率"
from course c
left join score sc on sc.c_id=c.c_id 
left join score sc2 on sc2.c_id=c.c_id 
left join score sc3 on sc3.c_id=c.c_id 
group by c.c_id

-- 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
-- 加@score是为了防止用union all 后打乱了顺序
select c1.s_id,c1.c_id,c1.c_name,@score:=c1.s_score,@i:=@i+1 from (select c.c_name,sc.* from course c 
left join score sc on sc.c_id=c.c_id
where c.c_id="01" order by sc.s_score desc) c1 ,
(select @i:=0) a
union all 
select c2.s_id,c2.c_id,c2.c_name,c2.s_score,@ii:=@ii+1 from (select c.c_name,sc.* from course c 
left join score sc on sc.c_id=c.c_id
where c.c_id="02" order by sc.s_score desc) c2 ,
(select @ii:=0) aa 
union all
select c3.s_id,c3.c_id,c3.c_name,c3.s_score,@iii:=@iii+1 from (select c.c_name,sc.* from course c 
left join score sc on sc.c_id=c.c_id
where c.c_id="03" order by sc.s_score desc) c3;
set @iii=0;


-- 20、查询学生的总成绩并进行排名
select st.s_id,st.s_name
,(case when sum(sc.s_score) is null then 0 else sum(sc.s_score) end)
 from student st
left join score sc on sc.s_id=st.s_id
group by st.s_id order by sum(sc.s_score) desc

-- 21、查询不同老师所教不同课程平均分从高到低显示 
select t.t_id,t.t_name,c.c_name,avg(sc.s_score) from teacher t 
left join course c on c.t_id=t.t_id 
left join score sc on sc.c_id =c.c_id
group by t.t_id
order by avg(sc.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 a.* from (
select st.*,c.c_id,c.c_name,sc.s_score from student st
left join score sc on sc.s_id=st.s_id
inner join course c on c.c_id =sc.c_id and c.c_id="01"
order by sc.s_score desc LIMIT 1,2 ) a
union all
select b.* from (
select st.*,c.c_id,c.c_name,sc.s_score from student st
left join score sc on sc.s_id=st.s_id
inner join course c on c.c_id =sc.c_id and c.c_id="02"
order by sc.s_score desc LIMIT 1,2) b
union all
select c.* from (
select st.*,c.c_id,c.c_name,sc.s_score from student st
left join score sc on sc.s_id=st.s_id
inner join course c on c.c_id =sc.c_id and c.c_id="03"
order by sc.s_score desc LIMIT 1,2) c

-- 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 c.c_id,c.c_name 
,((select count(1) from score sc where sc.c_id=c.c_id and sc.s_score<=100 and sc.s_score>80)/(select count(1) from score sc where sc.c_id=c.c_id )) "100-85"
,((select count(1) from score sc where sc.c_id=c.c_id and sc.s_score<=85 and sc.s_score>70)/(select count(1) from score sc where sc.c_id=c.c_id )) "85-70"
,((select count(1) from score sc where sc.c_id=c.c_id and sc.s_score<=70 and sc.s_score>60)/(select count(1) from score sc where sc.c_id=c.c_id )) "70-60"
,((select count(1) from score sc where sc.c_id=c.c_id and sc.s_score<=60 and sc.s_score>=0)/(select count(1) from score sc where sc.c_id=c.c_id )) "60-0"
from course c order by c.c_id

-- 24、查询学生平均成绩及其名次 
set @i=0;
select a.*,@i:=@i+1 from (
select st.s_id,st.s_name,round((case when avg(sc.s_score) is null then 0 else avg(sc.s_score) end),2) "平均分" from student st
left join score sc on sc.s_id=st.s_id
group by st.s_id order by sc.s_score desc) a

-- 25、查询各科成绩前三名的记录
select a.* from (
 select st.s_id,st.s_name,c.c_id,c.c_name,sc.s_score from student st
 left join score sc on sc.s_id=st.s_id
 inner join course c on c.c_id=sc.c_id and c.c_id='01'
 order by sc.s_score desc LIMIT 0,3) a
union all 
select b.* from (
 select st.s_id,st.s_name,c.c_id,c.c_name,sc.s_score from student st
 left join score sc on sc.s_id=st.s_id
 inner join course c on c.c_id=sc.c_id and c.c_id='02'
 order by sc.s_score desc LIMIT 0,3) b
union all
select c.* from (
 select st.s_id,st.s_name,c.c_id,c.c_name,sc.s_score from student st
 left join score sc on sc.s_id=st.s_id
 inner join course c on c.c_id=sc.c_id and c.c_id='03'
 order by sc.s_score desc LIMIT 0,3) c

-- 26、查询每门课程被选修的学生数 
select c.c_id,c.c_name,count(1) from course c 
left join score sc on sc.c_id=c.c_id
inner join student st on st.s_id=c.c_id
group by st.s_id

-- 27、查询出只有两门课程的全部学生的学号和姓名
select st.s_id,st.s_name from student st 
left join score sc on sc.s_id=st.s_id
inner join course c on c.c_id=sc.c_id 
group by st.s_id having count(1)=2

-- 28、查询男生、女生人数
select st.s_sex,count(1) from student st group by st.s_sex

-- 29、查询名字中含有"风"字的学生信息
select st.* from student st where st.s_name like "%风%";

-- 30、查询同名同性学生名单，并统计同名人数 
select st.*,count(1) from student st group by st.s_name,st.s_sex having count(1)>1

-- 31、查询1990年出生的学生名单
select st.* from student st where st.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.c_id,c.c_name,avg(sc.s_score) from course c
inner join score sc on sc.c_id=c.c_id 
group by c.c_id order by avg(sc.s_score) desc,c.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 st.s_id,st.s_name,avg(sc.s_score) from student st
left join score sc on sc.s_id=st.s_id
group by st.s_id having avg(sc.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 st.s_id,st.s_name,sc.s_score from student st
inner join score sc on sc.s_id=st.s_id and sc.s_score<60
inner join course c on c.c_id=sc.c_id and c.c_name ="数学" 

-- 35、查询所有学生的课程及分数情况；
select st.s_id,st.s_name,c.c_name,sc.s_score from student st
left join score sc on sc.s_id=st.s_id
left join course c on c.c_id =sc.c_id
order by st.s_id,c.c_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 st2.s_id,st2.s_name,c2.c_name,sc2.s_score from student st2
left join score sc2 on sc2.s_id=st2.s_id
left join course c2 on c2.c_id=sc2.c_id 
where st2.s_id in(
select st.s_id from student st 
left join score sc on sc.s_id=st.s_id 
group by st.s_id having min(sc.s_score)>=70)
order by s_id

-- 37、查询不及格的课程
select st.s_id,c.c_name,st.s_name,sc.s_score from student st
inner join score sc on sc.s_id=st.s_id and sc.s_score<60
inner join course c on c.c_id=sc.c_id 

-- 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 st.s_id,st.s_name,sc.s_score from student st
inner join score sc on sc.s_id=st.s_id and sc.c_id="01" and sc.s_score>=80

-- 39、求每门课程的学生人数
select c.c_id,c.c_name,count(1) from course c
inner join score sc on sc.c_id=c.c_id
group by c.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 st.*,c.c_name,sc.s_score,t.t_name from student st
inner join score sc on sc.s_id=st.s_id
inner join course c on c.c_id=sc.c_id 
inner join teacher t on t.t_id=c.t_id and t.t_name="张三"
order by sc.s_score desc
limit 0,1

-- 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 st.s_id,st.s_name,sc.c_id,sc.s_score from student st 
left join score sc on sc.s_id=st.s_id
left join course c on c.c_id=sc.c_id
where (
select count(1) from student st2 
left join score sc2 on sc2.s_id=st2.s_id
left join course c2 on c2.c_id=sc2.c_id
where sc.s_score=sc2.s_score and c.c_id!=c2.c_id 
)>1

-- 42、查询每门功成绩最好的前两名 
select a.* from (select st.s_id,st.s_name,c.c_name,sc.s_score from student st
left join score sc on sc.s_id=st.s_id
inner join course c on c.c_id=sc.c_id and c.c_id="01"
order by sc.s_score desc limit 0,2) a
union all
select b.* from (select st.s_id,st.s_name,c.c_name,sc.s_score from student st
left join score sc on sc.s_id=st.s_id
inner join course c on c.c_id=sc.c_id and c.c_id="02"
order by sc.s_score desc limit 0,2) b
union all
select c.* from (select st.s_id,st.s_name,c.c_name,sc.s_score from student st
left join score sc on sc.s_id=st.s_id
inner join course c on c.c_id=sc.c_id and c.c_id="03"
order by sc.s_score desc limit 0,2) c
 
-- 借鉴(更准确,漂亮):
 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 sc.c_id,count(1) from score sc
left join course c on c.c_id=sc.c_id
group by c.c_id having count(1)>5
order by count(1) desc,sc.c_id asc

-- 44、检索至少选修两门课程的学生学号 
select st.s_id from student st 
left join score sc on sc.s_id=st.s_id
group by st.s_id having count(1)>=2

-- 45、查询选修了全部课程的学生信息
select st.* from student st 
left join score sc on sc.s_id=st.s_id
group by st.s_id having count(1)=(select count(1) from course)

-- 46、查询各学生的年龄
 select st.*,timestampdiff(year,st.s_birth,now()) from student st

-- 47、查询本周过生日的学生
 -- 此处可能有问题,week函数取的为当前年的第几周,2017-12-12是第50周而2018-12-12是第49周,可以取月份,day,星期几(%w),
 -- 再判断本周是否会持续到下一个月进行判断,太麻烦,不会写
select st.* from student st 
where week(now())=week(date_format(st.s_birth,'%Y%m%d'))

-- 48、查询下周过生日的学生
select st.* from student st 
where week(now())+1=week(date_format(st.s_birth,'%Y%m%d'))

-- 49、查询本月过生日的学生
select st.* from student st 
where month(now())=month(date_format(st.s_birth,'%Y%m%d'))

-- 50、查询下月过生日的学生
 -- 注意:当 当前月为12时,用month(now())+1为13而不是1,可用timestampadd()函数或mod取模
select st.* from student st 
where month(timestampadd(month,1,now()))=month(date_format(st.s_birth,'%Y%m%d'))
-- 或
select st.* from student st where (month(now()) + 1) mod 12 = month(date_format(st.s_birth,'%Y%m%d'))

到此这篇关于Mysql Sql 语句练习题（50道）的文章就介绍到这了,更多相关Mysql练习题内容请搜索易盾网络以前的文章或继续浏览下面的相关文章希望大家以后多多支持易盾网络！