Mysql:查找行，其中时间戳差异小于x [英] Mysql: Find rows, where timestamp difference is less than x

问题描述

我的表结构如下:

Field       Type
id              int(11)      AI
user            varchar(64)
date            timestamp
key             int(11)

我需要做的是查找行(从给定的日期开始)，其中两个连续行之间的时间差(最近的时间戳记(给定用户))小于1300.

What I need to do, is find rows (from given day), where difference between two successive rows (closest timestamp (for given user) is less than 1300.

有人告诉我，使用这样的查询:

I was told, to use query like this:

select t.*
from (select t.*, @nextdate as nextdate, @nextdate := date
      from my_table t
      order by date desc
     ) t
where t.nextdate - t.date < 1300;

但是它似乎没有用.谁能帮助我解决任务?

But it didn't seem to work. Can anyone help me solve my task?

推荐答案

尝试一下:-

select t1.user, t1.date d1,t2.date d2 ,t1.date-t2.date
  from  (select @val:=@val+1 rowid,user,  date
           from mytable,(select @val:=0) a
       order by user,date) t1,

        (select @val1:=@val1+1 rowid,user,  date
           from mytable,(select @val1:=1) b
       order by user,date) t2

 where t1.rowid = t2.rowid
   and t1.user = t2.user
   and t1.date-t2.date < 1300;

请参见演示

这篇关于Mysql:查找行，其中时间戳差异小于x的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！