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17年10月份数据)
select * from exam where date_format(starttime,'%Y-%m')='2017-10'
select * from exam where date_format(starttime,'%Y-%m')=date_format('2017-10-05','%Y-%m')
【当你用心写完每一篇博客之后,你会发现它比你用代码实现功能更有成就感!】