mysql todays()函数的用法

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  1、利用to_days函数查询今天的数据:

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

  to_days函数:返回从0000年(公元1年)至当前日期的总天数。
  2、昨天
  SELECT * FROM 表名 WHERE TO_DAYS( NOW( ) ) – TO_DAYS( 时间字段名) <= 1
  3.7天
  SELECT * FROM 表名 where DATE_SUB(CURDATE(), INTERVAL 7 DAY) <= date(时间字段名)
  4.近30天
  SELECT * FROM 表名 where DATE_SUB(CURDATE(), INTERVAL 30 DAY) <= date(时间字段名)
  5.本月
  SELECT * FROM 表名 WHERE DATE_FORMAT( 时间字段名, ‘%Y%m' ) = DATE_FORMAT( CURDATE( ) , ‘%Y%m' )
  6.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`
posted @ 2020-12-18 14:11  优优鱼  阅读(6351)  评论(0编辑  收藏  举报