mysql数据库要按当天、昨天、前七日、近三十天、季度、年查询

今天笔者做一个销售系统查询，要按当天、昨天、前七日、近三十天、季度、年查询，我用的是php+mysql制作，常用的时间查询条件mysql语句，笔者在本地测试并且使用了，没有错！分享给大家了

查询今天 sql语句
select * from 表名 where to_days(时间字段名) = to_days(now());  

 

查询昨天 sql语句

SELECT * FROM 表名 WHERE TO_DAYS(NOW())-TO_DAYS(`时间字段名`) = 1 

查询7天 sql语句

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

 

查询近30天 sql语句

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

查询本月 sql语句

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

 

查询上一月 sql语句

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

同时，再附上 一个 mysql官方的相关document

#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();

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

where   date(regdate)   =   curdate();

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

SELECT date( c_instime ) ,curdate( )
FROM `t_score`
WHERE 1
LIMIT 0 , 30