MySQL中, 如何查询某一天, 某一月, 某一年的数据.

今天

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

昨天

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

近7天

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();

查询某个月的数据(查询18年11月份数据)

select * from exam where date_format(starttime,'%Y-%m')='2018-11'

select * from exam where date_format(starttime,'%Y-%m')=date_format('2018-11-11','%Y-%m')
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  1. #1

    正好需要用到,哈哈

    匿名2个月前 (04-15)回复