Mysql_45道练习题

我使用的Mysql版本是5.7.19。答案可能会因版本会有少许出入。如果答案有问题欢迎在评论区指出并给出你的答案。

练习数据

数据表

--1.学生表 Student(SId,Sname,Sage,Ssex)
    SId 学生编号,Sname 学生姓名,Sage 出生年月,Ssex 学生性别
--2.课程表 Course(CId,Cname,TId)
    CId 课程编号,Cname 课程名称,TId 教师编号
--3.教师表 Teacher(TId,Tname)
    TId 教师编号,Tname 教师姓名
--4.成绩表 SC(SId,CId,score)
    SId 学生编号,CId 课程编号,score 分数
创建测试数据

学生表 Student：
导入数据方法：将以下 mysql 语句，完整复制到 workbench 语句窗口（或者是 mysql 的黑窗口），然后运行即可导入，不需要另外创建表，下面表的操作一样。这些语句第一条是创建表（create table），后面都是插入数据到表中（insert into table ）。
create table Student(SId varchar(10),Sname varchar(10),Sage datetime,Ssex varchar(10));
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('09' , '张三' , '2017-12-20' , '女');
insert into Student values('10' , '李四' , '2017-12-25' , '女');
insert into Student values('11' , '李四' , '2017-12-30' , '女');
insert into Student values('12' , '赵六' , '2017-01-01' , '女');
insert into Student values('13' , '孙七' , '2018-01-01' , '女');

科目表 Course
create table Course(CId varchar(10),Cname nvarchar(10),TId varchar(10));
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 Teacher
create table Teacher(TId varchar(10),Tname varchar(10));
insert into Teacher values('01' , '张三');
insert into Teacher values('02' , '李四');
insert into Teacher values('03' , '王五');

成绩表 SC
create table SC(SId varchar(10),CId varchar(10),score decimal(18,1));
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);

练习题目

1.查询" 01 "课程比" 02 "课程成绩高的学生的信息及课程分数
1.1 查询同时存在" 01 "课程和" 02 "课程的情况
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 ) 1.3 查询不存在" 01 "课程但存在" 02 "课程的情况
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
3.查询在 SC 表存在成绩的学生信息
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 ) 4.1 查有成绩的学生信息
5.查询「李」姓老师的数量
6.查询学过「张三」老师授课的同学的信息
7.查询没有学全所有课程的同学的信息
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
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
10.查询没学过"张三"老师讲授的任一门课程的学生姓名
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
12.检索" 01 "课程分数小于 60，按分数降序排列的学生信息
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
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
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 15.1 按各科成绩进行排序，并显示排名， Score 重复时合并名次
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 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
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
18.查询各科成绩前三名的记录
19.查询每门课程被选修的学生数
20.查询出只选修两门课程的学生学号和姓名
21.查询男生、女生人数
22.查询名字中含有「风」字的学生信息
23.查询同名同性学生名单，并统计同名人数
24.查询 1990 年出生的学生名单
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
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
27.查询课程名称为「数学」，且分数低于 60 的学生姓名和分数
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）
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
30.查询不及格的课程
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
32.求每门课程的学生人数
33.假设成绩不重复，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
34.假设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
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
36.查询每门功成绩最好的前两名
37.统计每门课程的学生选修人数（超过 5 人的课程才统计）。
38.检索至少选修两门课程的学生学号
39.查询选修了全部课程的学生信息
40.查询各学生的年龄，只按年份来算
41.按照出生日期来算，当前月日 < 出生年月的月日则，年龄减一
42.查询本周过生日的学生
43.查询下周过生日的学生
44.查询本月过生日的学生
45.查询下月过生日的学生

参考答案：

部分题目讲解视频：https://pan.baidu.com/s/1SfBQAUbfBkDWh2-pGvrayg 提取码：2hlq （如果有需要讲解的题目可以在评论处留言）
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 * 
from sc a 
left join student d
    on a.sid=d.sid 
inner join sc b 
    on a.sid=b.sid and a.cid='01' and b.cid='02'
where a.score>b.score;

1.1 查询同时存在" 01 "课程和" 02 "课程的情况

select * 
from sc a 
inner join sc b 
    on a.sid=b.sid and a.cid='01' and b.cid='02';

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 )

select * 
from sc a 
left join sc b 
    on a.sid=b.sid  and b.cid='02'
where a.cid='01';

1.3 查询不存在" 01 "课程但存在" 02 "课程的情况

select * 
from sc a 
where sid not in(select sid from sc where cid='01')
and cid='02';

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
问题：为什么这里的group by 的having 条件查询，可以直接使用 s_avg 别名？

select 
a.sid,
d.sname,
avg(score) as s_avg
from sc a
left join student d
on a.sid=d.sid
group by a.sid
having s_avg>60;

查询在 SC 表存在成绩的学生信息

select 
    b.*
from sc a
left join student b
on a.sid=b.sid
group by a.sid;

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)

select 
a.sid,
a.sname,
count(b.cid) as cons,
sum(b.score) as cours
from student a
left join sc b
on a.sid=b.sid
group by a.sid;

4.1 查有成绩的学生信息

select
*
from student a
where a.sid in (select sid from sc group by sid);

或者

select *
from student
where EXISTS(select * from sc where student.SId=sc.SId)

5.查询「李」姓老师的数量

select
count(1)
from teacher
where tname like '李%';

6.查询学过「张三」老师授课的同学的信息

select 
b.*
from sc a 
left join student b 
on a.sid=b.sid 
where cid in (select cid from course 
where tid=(select tid from teacher 
where tname='张三')
) 
group by sid;

或者

select student.*
from teacher  ,course  ,student,sc
where teacher.Tname='张三'
and   teacher.TId=course.TId
and   course.CId=sc.CId
and   sc.SId=student.SId

7.查询没有学全所有课程的同学的信息
用学习信息关联到成绩sc 表得到每个人的课程信息，然后分组计数就知道是不是满足全部学习了。这里需要成绩表一门功课只有一条记录

select 
    * 
from student a 
left join sc b 
    on a.sid=b.sid 
group by a.sid 
    having count(b.cid)<(select count(cid) 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

select distinct b.* 
from sc a 
left join student b 
on a.sid=b.sid 
where cid in (select cid from sc where sid='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

select b.*
from (select * from sc where sid not in (select sid from sc where cid not in (select cid from sc where sid='01')) and sid!='01') a
left join student b
on a.sid=b.sid
group by a.sid
having count(cid)=(select count(cid) from sc where sid='01');

10.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 a
where a.sid not in(
select sid
from sc a
left join course c
on a.cid=c.cid
inner join teacher d
on c.tid=d.tid and d.tname='张三' )

或者

select *
from student 
where student.SId not in 
(
select student.SId
from student left join sc on student.SId=sc.SId 
where EXISTS 
(select *
from teacher ,course
where teacher.Tname='张三'
and   teacher.TId=course.TId
and     course.CId=sc.CId))

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

select *,avg(score) as avg_sco
from sc a
left join student b
on a.sid=b.sid
where a.score<60
group by a.sid having count(cid)>=2;

12、检索" 01 "课程分数小于 60，按分数降序排列的学生信息

select *
from sc a
left join student b
on a.sid=b.sid
where a.cid='01' and a.score<60
order by a.score desc;

或者

select student.*
from student,sc
where sc.CId ='01'
and   sc.score<60
and   student.SId=sc.SId

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 c.sid,c.cid,c.score,d.avg_sco
from 
(select 
a.sid,b.cid,b.score
from student a
left join sc b
on a.sid=b.sid ) c
left join 
 (select 
sid,avg(a.score) as avg_sco
from sc a
group by a.sid) d
on c.sid=d.sid 
order by avg_sco desc;

没有关联上 student 表，不过关联一下也可以

select 
sc.SId,sc.CId,sc.score,t1.avgscore 
from  sc left join (select sc.SId,avg(sc.score) as avgscore 
from sc 
GROUP BY sc.SId) as t1 on sc.SId =t1.SId 
ORDER BY t1.avgscore DESC
select sc.CId , case when @fontscore=score then @curRank when @fontscore:=score then @curRank:=@curRank+1  end as rank,sc.score
from (select @curRank:=0 ,@fontage:=null) as t ,sc
ORDER BY sc.score desc
select sc.CId ,max(sc.score)as 最高分,min(sc.score)as 最低分,AVG(sc.score)as 平均分,count(*)as 选修人数,sum(case when sc.score>=60 then 1 else 0 end )/count(*)as 及格率,sum(case when sc.score>=70 and sc.score<80 then 1 else 0 end )/count(*)as 中等率,sum(case when sc.score>=80 and sc.score<90 and sc.score<80 then 1 else 0 end )/count(*)as 优良率,sum(case when sc.score>=90 then 1 else 0 end )/count(*)as 优秀率 
from sc
GROUP BY sc.CId
ORDER BY count(*)DESC,sc.CId asc

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

select 
cid,
max(score) as max_sc,
min(score) as min_sc,
avg(score) as avg_sc,
count(1) as cons,
sum(if(score>=60,1,0))/count(1) as jige,
sum(if(70<=score and score<=80,1,0))/count(1) as zd,
sum(if(80<=score and score<=90,1,0))/count(1) as yl,
sum(if(90<=score ,1,0))/count(1) as yx
from sc group by cid;

或者用 case when ，也可以

select 
sc.CId ,
max(sc.score)as 最高分,
min(sc.score)as 最低分,
AVG(sc.score)as 平均分,
count(*)as 选修人数,
sum(case when sc.score>=60 then 1 else 0 end )/count(*)as 及格率,
sum(case when sc.score>=70 and sc.score<80 then 1 else 0 end )/count(*)as 中等率,
sum(case when sc.score>=80 and sc.score<90 then 1 else 0 end )/count(*)as 优良率,
sum(case when sc.score>=90 then 1 else 0 end )/count(*)as 优秀率 
from sc
GROUP BY sc.CId
ORDER BY count(*)DESC,sc.CId asc

15、按各科成绩进行排序，并显示排名， Score 重复时也继续排名

select 
    sid,cid,score,
    @rank:=@rank+1 as rn 
from sc ,(select @rank:=0) as t order by score desc;

15.1 按各科成绩进行排序，并显示排名， Score 重复时合并名次

select 
*,
case when (@sco=score) then @rank else @rank:=@rank+1 end as rn,
@sco:=score  -- 保存上一次的分数
 from sc ,(select @rank:=0,@sco:=null) as t order by score desc

或者

select 
sc.CId , 
case when @fontscore=score then @curRank 
    when @fontscore:=score then @curRank:=@curRank+1  
    end as rank,sc.score
from (select @curRank:=0 ,@fontage:=null) as t ,sc
ORDER BY sc.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

select 
s.*,
case when @sco=scos then '' else @rank:=@rank+1 end as rn ,
@sco:=scos
from 
(select 
sid,sum(score) as scos 
from sc group by sid order by scos desc) s,
(select @rank:=0,@sco:=null) as t

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
这个空缺的定义比较模糊，看看就好。

select t1.*, 
    case when @fontscore=t1.sumscore then @currank  
    when @fontscore:=t1.sumscore  then @currank:=@currank+1  
    end as rank
from (select sc.SId, sum(score) as sumscore
from sc
GROUP BY sc.SId 
ORDER BY sum(score) desc) as t1,(select @currank:=0,@fontscore:=null) as t

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

select 
*,
sum(case when 0<=score and score <=60 then 1 else 0 end )/count(1) as '[0,60]',
sum(case when 60<score and score <=70 then 1 else 0 end )/count(1) as '[60,70]',
sum(case when 70<score and score <=85 then 1 else 0 end )/count(1) as '[70,85]',
sum(case when 85<score and score <=100 then 1 else 0 end )/count(1) as '[85,100]'
from sc group by cid

或者

select course.CId,course.Cname,t1.*
from course LEFT JOIN (
select sc.CId,CONCAT(sum(case when sc.score>=85 and sc.score<=100 then 1 else 0 end )/count(*)*100,'%') as '[85-100]',
CONCAT(sum(case when sc.score>=70 and sc.score<85 then 1 else 0 end )/count(*)*100,'%') as '[70-85)',
CONCAT(sum(case when sc.score>=60 and sc.score<70 then 1 else 0 end )/count(*)*100,'%') as '[60-70)',
CONCAT(sum(case when sc.score>=0 and sc.score<60 then 1 else 0 end )/count(*)*100,'%') as '[0-60)'
from sc
GROUP BY sc.CId) as t1 on course.CId=t1.CId

18、查询各科成绩前三名的记录
思路：前三名转化为若大于此成绩的数量少于3即为前三名。

select 
*
from sc a
where (select count(1) from sc b where a.cid=b.cid and b.score>a.score)<3;

或者

select *
from sc  
where  (select count(*) from sc as a where sc.CId =a.CId and  sc.score <a.score )<3
ORDER BY CId asc,sc.score desc

19、查询每门课程被选修的学生数

select 
cid,
count(1) as cons
from sc group by cid;

20、查询出只选修两门课程的学生学号和姓名

select student.SId,student.Sname
from sc,student
where student.SId=sc.SId  
GROUP BY sc.SId
HAVING count(*)=2

21.查询男生、女生人数

select count(1) as cons,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，并统计同名同姓人数

select a.*,count(1) as cons 
from student a 
inner join student b 
on a.sname=b.sname and a.ssex=b.ssex and a.sid!=b.sid;

或者

select *
from student LEFT JOIN (select Sname,Ssex,COUNT(*)同名人数 from Student group by Sname,Ssex) as t1
on student.Sname =t1.Sname and student.Ssex=t1.Ssex
where t1.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 cid,avg(score) as avg_sco 
from sc 
group by cid 
order by avg_sco desc,cid 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

select * from student a 
inner join 
(select sid,avg(score) as avg_sco 
from sc 
group by sid 
having avg_sco>=85) b 
on a.sid=b.sid;

27、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.sname,
b.cid,
b.score
from student a
inner join sc b
on a.sid=b.sid
where cid=(select cid from course where cname='数学')
and b.score<60;

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）

select * from student a left join sc b on a.sid=b.sid;

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

select *
from sc a
left join student b
on a.sid=b.sid
left join course c
on a.cid=c.cid
where a.score>70

30.查询存在不及格的课程

select 
*
from sc a
left join course b
on a.cid=b.cid
where a.score<60 
group by b.cid,b.cname;

或者

select DISTINCT sc.CId
from sc
where sc.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

select student.SId,student.Sname
from student ,sc
where sc.CId='01'
and  student.SId=sc.SId
and  sc.score>80

32、求每门课程的学生人数

select sc.CId,count(*) as 学生人数
from sc
GROUP BY sc.CId

33、假设成绩不重复，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

select 
*
from sc a
inner join student b
on a.sid=b.sid
inner join course c
on a.cid=c.cid 
inner join teacher d
on c.tid=d.tid and d.tname='张三'
order by a.score desc limit 1;

或者

select student.*,sc.score
from student ,course ,teacher ,sc
where course.CId=sc.CId
and course.TId=teacher.TId
and teacher.Tname='张三'
and student.SId =sc.SId
LIMIT 1

34、假设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
成绩重复，也就是有可能第一名是同分数的，也都要查询出来，这种也可以通过 count(*)<1 来实
修改为得到数据后，对子查询表进行排名编号，这样排名是对的

select 
a.*,
case when @score=score then @rank
when @score:=score then @rank:=@rank+1 end as rn
from 
(select 
a.sid,
a.score,
c.cid,
d.tname
from sc a
left join course c
on a.cid=c.cid 
left join teacher d
on c.tid=d.tid 
where d.tname='张三') a
,(select @score:=null,@rank:=0) as t
order by a.score desc;

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

select 
*
from sc a
inner join sc b
on a.sid=b.sid
where a.cid!=b.cid and a.score=b.score
group by a.sid ,a.cid

或者

select *
from sc as t1
where exists(select * from sc as t2 
        where t1.SId=t2.SId 
        and t1.CId!=t2.CId 
        and t1.score =t2.score )

36.查询每门功成绩最好的前两名

select 
*
from sc a
where (select count(1) from sc b where a.cid=b.cid and b.score>a.score)<=1;

或者

select *
from sc as t1
where (select count(*) from sc as t2 
    where t1.CId=t2.CId and t2.score >t1.score)<2
ORDER BY t1.CId

37.统计每门课程的学生选修人数（超过 5 人的课程才统计）

select
cid,
count(1) as cons
from sc
group by cid
having cons>5;

或者

select sc.CId as 课程编号,count(*) as 选修人数
from sc 
GROUP BY sc.CId
HAVING count(*)>5

38.检索至少选修两门课程的学生学号

select 
sid,
count(1) as cons
from sc
group by sid
having cons>=2;

或者

select DISTINCT t1.SId
from sc as t1 
where (select count(* )from sc where t1.SId=sc.SId)>=2

39、查询选修了全部课程的学生信息

select 
*
from student a
where (select count(1) from sc b where a.sid=b.sid)
    =(select count(1) 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、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 
    *,
    case when substr(sage,6,5)<substr(now(),6,5) 
            then year(now())-year(sage) 
        when substr(sage,6,5)>=substr(now(),6,5) 
            then year(now())-year(sage) -1 end as age
from student;

或者
这里 timestampdiff 会用年月日去计算 年 的相隔时间，
如果相差1.9年则为1年，所以实际上是已经相减了的，正好用来计算生日

select 
    student.SId as 学生编号,
    student.Sname  as  学生姓名,
    TIMESTAMPDIFF(YEAR,student.Sage,CURDATE()) as 学生年龄
from student

42.查询本周过生日的学生

select 
*,
substr(YEARWEEK(student.Sage),5,2) as birth_week
substr(YEARWEEK(CURDATE()),5,2) as now_week
from student 
where substr(YEARWEEK(student.Sage),5,2)=substr(YEARWEEK(CURDATE()),5,2);

43、查询下周过生日的学生

select 
    *,
    substr(YEARWEEK(student.Sage),5,2) as birth_week,
    substr(YEARWEEK(CURDATE()),5,2) as now_week 
from student 
where substr(YEARWEEK(student.Sage),5,2)=
        substr(YEARWEEK(CURDATE()),5,2)+1;

多种 week 的获取方法

select *,
    week(sage)  as sweek,
    yearweek(sage) as yweek,
    EXTRACT(week FROM student.Sage) as eweek,
    EXTRACT(week FROM now()) as nweek
from student 
where EXTRACT(week FROM Sage)=EXTRACT(week FROM now())+1

44.查询本月过生日的学生

select 
*,
month(sage) as birth_month,
month(now()) as now_month
from student
where month(sage)=month(now())

或者

select *
from student 
where EXTRACT(MONTH FROM student.Sage)=
    EXTRACT(MONTH FROM CURDATE())

45.查询下月过生日的学生

select 
*,
month(sage) as birth_month,
month(now()) as now_month
from student
where month(sage)=month(now())+1

或者

select *
from student 
where EXTRACT(MONTH FROM student.Sage)=
    EXTRACT(MONTH FROM DATE_ADD(CURDATE(),INTERVAL 1 MONTH))