共计 2016 个字符，预计需要花费 6 分钟才能阅读完成。

SELECT * FROM 表名 where DATE_SUB(CURDATE(), INTERVAL 7 DAY) = date(时间字段名)

近 30 天

SELECT * FROM 表名 where DATE_SUB(CURDATE(), INTERVAL 30 DAY) = date(时间字段名)

本月

SELECT * FROM 表名 WHERE DATE_FORMAT(时间字段名, '%Y%m') = DATE_FORMAT(CURDATE() , '%Y%m' )

上一月

SELECT * FROM 表名 WHERE PERIOD_DIFF(date_format( now() , '%Y%m' ) , date_format(时间字段名, '%Y%m') ) =1

查询本季度数据

select * from `ht_invoice_information` where QUARTER(create_date)=QUARTER(now());

查询上季度数据

select * from `ht_invoice_information` where QUARTER(create_date)=QUARTER(DATE_SUB(now(),interval 1 QUARTER));

查询本年数据

select * from `ht_invoice_information` where YEAR(create_date)=YEAR(NOW());

查询上年数据

select * from `ht_invoice_information` where year(create_date)=year(date_sub(now(),interval 1 year));

查询当前这周的数据

SELECT name,submittime FROM enterprise WHERE YEARWEEK(date_format(submittime,'%Y-%m-%d')) = YEARWEEK(now());

查询上周的数据

SELECT name,submittime FROM enterprise WHERE YEARWEEK(date_format(submittime,'%Y-%m-%d')) = YEARWEEK(now())-1;

查询上个月的数据

select name,submittime from enterprise where date_format(submittime,'%Y-%m')=date_format(DATE_SUB(curdate(), INTERVAL 1 MONTH),'%Y-%m')
select * from user where DATE_FORMAT(pudate,'%Y%m') = DATE_FORMAT(CURDATE(),'%Y%m') ; 
select * from user where WEEKOFYEAR(FROM_UNIXTIME(pudate,'%y-%m-%d')) = WEEKOFYEAR(now()) 
select * from user where MONTH(FROM_UNIXTIME(pudate,'%y-%m-%d')) = MONTH(now()) 
select * from user where YEAR(FROM_UNIXTIME(pudate,'%y-%m-%d')) = YEAR(now()) and MONTH(FROM_UNIXTIME(pudate,'%y-%m-%d')) = MONTH(now()) 
select * from user where pudate between 上月最后一天 and 下月第一天 

查询当前月份的数据

select name,submittime from enterprise where date_format(submittime,'%Y-%m')=date_format(now(),'%Y-%m')

查询距离当前现在 6 个月的数据

select name,submittime from enterprise where submittime between date_sub(now(),interval 6 month) and now();

PS: 下面看下 mysql 如何查询当天信息？

原来不是太熟悉 SQL 查询语句，什么都是用到了再去查去找，还好网络提供给我们很多支持。今天又用到了一个语句，一时间真想不出怎么解决，到网上看了看，感觉就有一个，怎么那么简单啊。需要积累的东西真是太多了。

今天就把我这个简单的问题记录下来吧! 算是一个积累：

mysql 查询当天的所有信息：

select * from test where year(regdate)=year(now()) and month(regdate)=month(now()) and day(regdate)=day(now())

这个有一些繁琐，还有简单的写法：

select * from table where date(regdate) = curdate();

date() 函数获取日期部分, 扔掉时间部分，然后与当前日期比较即可

向 AI 问一下细节

丸趣 TV 网 – 提供最优质的资源集合！