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mysql根据日期查询!!

使用mysql的日期函数吧: select * from A where DATE(data) = '2012-07-09'; 这个可以准确的查询到7.9号这天的数据,而且效率很高。

select * from job where date>=2008-07-02 and date

SELECT DATE_FORMAT(time,'%Y-%m-%d') as day, sum(case when amount>0 then amount when amount=0 then 0 end) as amount1 from table where time>='2014-11-01' group by day; 我没有测试。time表示日期,amount表示数量。查询11月后每天成交数量

你的要求是 '2010-7-12 11:18:54' 到 '2010-7-12 11:22:20' 之间的 显然2010-7-12 11:18:46 和 2010-7-12 11:22:22不在二者之间,为什么要包含在结果集内? 我也附议楼上三位 select sum(csize) from table where cdtime between '2010-7-12 11:...

使用date_format将表中datetime字段的值转换成“年月日”格式的字符串即可 语句这样写: select * from 表名 where date_format(日期字段,'%Y-%m-%d') = ‘2012-1-1’; 记得采纳。

1. sql 中字段不要加‘ ’ =>select *, caozuo as caozuo from mj_searched (多此一举吗? 有*了 还艺单独查caozuo) 2. select *, caozuo as caozuo from mj_searched where publishedtime >= '2013-02-12' AND publishedtime

MySQL数据库中year()函数是求某个特定日期中的年份,代码如下: select '2015-08-11' as date,year('2015-08-11') as year; 确定一个日期是一年中的第几个季度,可以用QUARTER()函数实现,代码如下: SELECT '2015-08-11' AS DATE,QUARTER('2015...

类似于:select * from tableA where addtime>='2015-10-20 00:00:00' and addtime

可以用分组汇总求极值来实现这个需求,请参考下列写法: select `name`,test,max(`time`) as lastTime from tblname where `name`='test' group by `name`,test; 如果要取整条记录的会麻烦一些,必须有唯一标识列,如主键、自增id等才可以实现,...

SELECTaa.*, bb.VALUEFROM(SELECTNAME,max(time) AS timeFROMmy_testGROUP BYNAME) aaLEFT JOIN (SELECTNAME,time,sum(VALUE) ASVALUEFROMmy_testGROUP BYNAME,time) bb ON aa. NAME = bb. NAMEAND aa.time = bb.timeps:写得有点复杂 (my_test ...

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