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mysql 时间间隔查询

select count(*) from sometable where datetimecolumn>=UNIX_TIMESTAMP('2010-03-01 00:00:00') and datetimecolumn

你表中的数据,根据你的题目,应该是全部显示才对。是小于5秒还是5分钟? 下面这个是 两条数据间小于5秒的记录的查询语句 select *,timestampdiff(second,t1.time,(select time from table3 where t1.idid order by id desc limit 1),t1.time ) ...

select * from table where 时间字段 between '2010-7-12 11:18:54' and '2010-7-12 11:22:20'

应该先算出 从开始到结束 的时间和,然后判断是否大于24 例如 22 -10 = 10 程序里面取整数或者小数 22.5-10.5 =10 22.3 - 10.2 = 10.1 个人认为是这样,祝你顺利!

一楼的写错,二楼的乱写, 假设存在一主键列id select * from 表 where id in(select id from 表 group by id haing count(数据列)=3) 这句就行了~

mysql要实现定时执行sql语句就要用到Event 具体操作如下: 先看看看event 事件是否开启 show variables like '%sche%'; 如没开启,则开启。需要数据库超级权限 set global event_scheduler =1; 创建存储过程 update_a (注:就是你要执行的sql语...

TIMESTAMPDIFF(interval,datetime_expr1,datetime_expr2)返回日期或日期时间表达式datetime_expr1 和datetime_expr2the 之间的整数差。其结果的单位由interval 参数给出。该参数必须是以下值的其中一个:FRAC_SECOND 表示间隔是毫秒SECOND 秒MIN...

select timestampdiff(minute,"2016-08-09 11:15:16","2016-08-09 11:55:16");

时间段统计,可以采用 hour(subscribe_time) 取出小时然后分层。思路: select uid ,CASE WHEN HOUR(subscribe_time) BETWEEN 0 AND 1 THEN '00:00:00' WHEN HOUR(subscribe_time) BETWEEN 2 AND 3 THEN '02:00:00' ... ELSE '23:00:00' END -- ...

不考虑 “天”的因素: select if(month(d2)-month(d1)>0, concat(year(d2)-year(d1),'年',month(d2)-month(d1),'月'), concat(year(d2)-year(d1)-1,'年',month(d2)-month(d1)+12,'月')) as r from xxxxx ---------------------------------------...

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