Question

在下表示例 - 表A中，我们有四个不同ID 1,2,3,4的条目，分别具有相应的状态和时间。我想找到＆＃34; ID＆＃34;花了最多的时间来改变＆＃34;状态＆＃34;从开始到完成。在下面的示例中，它是ID = 4.我想运行查询并查找结果，我们目前在表中有大约百万条记录。如果有人提供有效的方法来检索这些数据，那将是非常好的。

表A

ID    Status         Date(YYYY-DD-MM HH:MM:SS)
1.    Started        2017-01-01 01:00:00
1.    Completed      2017-01-01 02:00:00
2.    Started        2017-10-02 03:00:00
2.    Completed      2017-10-02 05:00:00
3.    Started        2017-15-03 06:00:00
3.    Completed      2017-15-03 09:00:00
4.    Started        2017-22-04 10:00:00
4.    Completed      2017-22-04 15:00:00

谢谢！布鲁斯

Answer 1

我没有运行此sql，但它可以解决您的问题。

select  a.id, max(DATEDIFF(SECOND, a.date, b.date + 1)) from TableA as a 
join TableA as b on a.id = b.id
where a.status="started" and b.status="completed"

Answer 2

SELECT
    started.ID, timediff(completed.date, started.date) as elapsed_time
FROM TABLE_A as started
INNER JOIN TABLE_A as completed ON (completed.ID=started.ID AND completed.status='Completed')
WHERE started.status='Started'
ORDER BY elapsed_time desc

确保TABLE_A上有列ID，日期

的索引

Answer 3

您可以查询如下：

Select top 1 with ties Id from #yourDate y1
    join #yourDate y2
On y1.Id = y2.Id
    and y1.[STatus] = 'Started'
    and y2.[STatus] = 'Completed'
    order by Row_number() over(order by  datediff(mi,y1.[Date], y2.[date]) desc)

Answer 4

这是一种使用相关子查询的方法。在这种情况下，只需取消注释TOP 1即可获得ID 4。这是基于你的评论，只有1＆＃34;开始＆＃34;记录，但可以是多个＆＃34;完成＆＃34;每个ID的记录。

declare @TableA table (ID int, Status varchar(64), Date datetime)
insert into @TableA
values
(1,'Started','2017-01-01 01:00:00'),
(1,'Completed','2017-01-01 02:00:00'),
(2,'Started','2017-02-10 03:00:00'),
(2,'Completed','2017-02-10 05:00:00'),
(3,'Started','2017-03-15 06:00:00'),
(3,'Completed','2017-03-15 09:00:00'),
(4,'Started','2017-04-22 10:00:00'),
(4,'Completed','2017-04-22 15:00:00')


select --top 1
    s.ID
    ,datediff(minute,s.Date,e.EndDate) as TimeDifference
from @TableA s
inner join(
    select
        ID
        ,max(Date) as EndDate
    from @TableA
    where Status = 'Completed'
    group by ID) e on e.ID = s.ID
where
    s.Status = 'Started'
order by
    datediff(minute,s.Date,e.EndDate)  desc

<强>返回

+----+----------------+
| ID | TimeDifference |
+----+----------------+
|  4 |            300 |
|  3 |            180 |
|  2 |            120 |
|  1 |             60 |
+----+----------------+

Answer 5

如果你知道＆＃39;已经开始＆＃39;永远是每个ID最早的时间点，也是最后一个完成的ID＃39;您正在考虑的记录将始终是每个ID的最新时间点，以下应该对大量记录具有良好的性能：

SELECT TOP 1
      id
    , DATEDIFF(s, MIN([Date]), MAX([date])) AS Elapsed
FROM   @TableA
GROUP BY ID
ORDER BY DATEDIFF(s, MIN([Date]), MAX([date])) DESC

查找单个列的最大日期差异

5 个答案: