MySQL中, 如何查询某一天, 某一月, 某一年的数据.

转自：http://www.cnblogs.com/benefitworld/p/5832897.html，稍作修改。

今天

select * from 表名 where to_days(时间字段名) = to_days(now());

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

SELECT * FROM 表名 WHERE DATEDIFF(字段,NOW())=-1;
-- 同理，查询前天的就是-2

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

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

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

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

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

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

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

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

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

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

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

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

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