Question

Type  Status    date                       ID           Users
-------------------------------------------------------------
  2       1     6/1/2015 8:43:13.557 AM    1140914474   2B 4E
  1       1     6/1/2015 8:43:40.057 AM    1140914474   2B 4E
  2       1     6/1/2015 12:29:15.093 PM    655666666   2B 4E
  1       1     6/1/2015 12:32:43.067 PM    655666666   2B 4E
  2       1     6/1/2015 6:02:30.557 PM    1694543676   2B 4E
  1       1     6/1/2015 6:14:34.180 PM    1694543676   2B 4E

我需要在行类型2和类型1之间获取日期差异，并将所有时间总计为天。

Answer 1

与SQL Server 2005相同:-)此查询将返回您要聚合的所有数据

select 
    cast(b.date as date) day, 
    datediff(s, a.date, b.date) diff 
from 
    mytable a 
join 
    mytable b on a.ID = b.ID 
where 
    a.Type = 2 and b.Type = 1

将此查询用作CTE：

with result as 
( 
     select 
         cast(b.date as date) dt, 
         datediff(s, a.date, b.date) diff 
     from
         mytable a 
     join 
         mytable b on a.ID = b.ID 
     where 
         a.Type = 2 and b.Type = 1
)
select 
    dt, sum(diff) difftotal  
from 
    result  
group by 
    dt;

Answer 2

您的问题的查询将如下：

select 
  datea,
  sum(differenceSec) as Total
from (
select a.type,
  cast(a.date as date) datea,
  cast(b.date as date) dateb,
  DATEDIFF ( ss , a.date , b.date ) as differenceSec
from data a
  inner join data b
    on b.id = a.id
      and b.type = 1
  where a.type = 2 ) result
 group  by datea

查询的第一部分，按日分组结果并求它们。在子查询中，我们获得了类型2和类型1之间的区别。

差异以秒为单位，但您可以根据需要进行更改。

在这里您可以找到一个示例：http://sqlfiddle.com/#!3/600f9/24包含您的数据。

查询组的第一部分按日结果并求和。在子查询中，我们获得了类型2和类型1之间的区别。

SQL Server 2012中的自联接

2 个答案: