Question

我有一张桌子：

Ticket #      SubmitDate    Status
    101        1/1/2015     Closed
    102        1/1/2015     Closed
    103        1/1/2015     Closed
    104        1/2/2015      Open
    105        1/2/2015     Closed
    106        1/2/2015      Open
    107        1/2/2015      Open
    108        1/2/2015     Closed
    109        1/3/2015     Closed
    110        1/4/2015      Open
    111        1/5/2015     Closed
    112        1/5/2015     Closed
    113        1/5/2015      Open
    114        1/5/2015      Open
    115        1/5/2015      Open

我正在努力寻找仍然开放的最旧票。我知道MIN（SubmitDate）会得到我最早的SubmitDate，但我无法将查询放在一起。

新编辑：大家好。我正在修改这个问题，所以更清楚。我向任何认为令人困惑的人道歉。我很有压力要立即得到答案，所以我太匆忙地贴了它。

我正在尝试确定仍处于“打开”状态的最旧票证＃。我一切都错了：

以下是我失败的查询之一：

SELECT

MIN(SubmitDate) as [Date Submitted]
,(SELECT FROM TicketTable WHERE Status = 'Open' AND SubmitDate = MIN(SubmitDate)) as [Ticket #]
,Status as [Status]

FROM TicketTable

我想要达到的结果是：

Date Submitted  Ticket #    Status
1/2/2015          104        Open

我再次道歉。我从来没有打算简单地得到答案。我只是希望有一丝暗示。再次感谢所有人。

Answer 1

select top 1 * from table
where "status" = 'Open'
order by "SubmitDate" asc ;

在SQL Server 2012+版本中，您还可以使用（标准）offset - fetch语法替换top：

select * from table
where "status" = 'Open'
order by "SubmitDate" asc 
offset 0 rows
fetch first 1 row only ;

Answer 2

select top 1 * from Table where Status = 'Open' order by SubmitDate desc

Answer 3

在row_number()上使用order by和SubmitDate条款就行了。

select SubmitDate, [Ticket #], Status
  from (select t.*,
               row_number() over (order by SubmitDate) as rn
          from mytable t
         where t.Status = 'Open') x
 where x.rn = 1

Answer 4

如您所见，您的大部分答案都是这种格式。我个人认为这是最直接的前进，因为它说的是混淆嵌套查询和使用分钟。

SELECT TOP 1 * 
FROM 
    YourTable 
WHERE 
    Status = 'Open'
ORDER BY
    SubmitDate Desc

Answer 5

<强> SQL Fiddle Demo

SELECT *
FROM YourTable Y,
    (SELECT  MIN(SubmitDate) mDate
    FROM YourTable
    WHERE Status = 'Open') as T
WHERE Y.SubmitDate = T.mDate
  AND Status = 'Open'

Answer 6

我只需要在已经很棒的列表中添加另一个：

SELECT * 
FROM table
WHERE submitDate IN (SELECT min(SubmitDate) FROM table WHERE status = 'Open')

但就个人而言，我会选择SELECT TOP 1...路线的@ STB，因为这很可能是该组中表现最佳的。其余的将依赖于子查询，这意味着DB（包括这一个）两次击中表。

但是......只是想分享一下，有很多方法可以解决这个问题

如何找到未关闭的最旧票？

6 个答案: