SQL 按一个字段分组并按另一个字段排序

Question

桌子：

declare @Table table (    
 id int,    
 ticketid int,    
 sponsor int,    
 dev int,
 qa int,
 savedate datetime
)

insert into @Table values (1,100,22,0, 0, '2008-10-29 11:17:59.527')
insert into @Table values (2,100,5,0, 0, '2008-10-29 11:00:37.030')
insert into @Table values (3,101,22,0, 0, '2009-10-29 11:10:27.687')
insert into @Table values (5,101,44,0, 0, '2008-10-31 12:07:52.917')
insert into @Table values (6,101,32,0, 0, '2009-06-30 08:16:12.343')
insert into @Table values (7,101,44,0, 0, '2009-10-31 10:12:11.369')

我正在尝试选择前 1 个最大保存数据，其中 recordid 是某个记录，按赞助商分组。

我的进步：

select max(savedate)
from @Table
where ticketid = 101
group by sponsor

返回

2009-10-29 11:10:27.687
2009-06-30 08:16:12.343
2009-10-31 10:12:11.370

关闭，我的分组正确，但我想要前 1 个最近的日期。所以我这样做：

select top 1 max(savedate)
from @Table
where ticketid = 101
group by sponsor

返回

2009-10-29 11:10:27.687

哇哦，明白了，该休息一下了..等等..这不是你最近的约会！让我们尝试通过 savedate 来订购这些

select top 1 max(savedate)
from @Table
where ticketid = 101
group by sponsor
order by savedate desc

哦不！可怕的：

“列“@Table.savedate”无效 在 ORDER BY 子句中，因为它是 不包含在任一聚合中 函数或 GROUP BY 子句。”

但 saveate 已聚合在选择列表中！我该怎么做？

Answer 1

在输入这个东西并确保我没有遗漏任何细节之后，我在最后想出了答案。我想无论如何我都会添加它，以便我或其他任何人稍后可以找到它，如果他们像我在这种情况下一样短视。

select top 1 max(savedate) as date
from @Table
where ticketid = 101
group by sponsor
order by date desc

max(savedate) 和 savedate 不一样！别名然后引用聚合工作完美：

2009-10-31 10:12:11.370

希望这对某人有所帮助。

Answer 2

select top 1 max(savedate)
from @Table
where ticketid = 101
group by sponsor
order by max(savedate) desc

Answer 3

您可能希望连同MAX(savedate) 一起退回赞助商。

如果你没有注意到这个查询：

SELECT  MAX(savedate)
FROM    @Table
WHERE   ticketid = 101

与您的解决方案完全相同

select top 1 max(savedate) as date
from @Table
where ticketid = 101
group by sponsor
order by date desc

，同时效率更高。

或者，您可以执行以下操作：

SELECT  TOP 1 sponsor, savedate
FROM    @Table
WHERE   ticketid = 101
ORDER BY
        savedate DESC

更新：

此查询将返回任务最后分配给当前赞助商的日期：

declare @Table table (    
 id int PRIMARY KEY,    
 ticketid int,    
 sponsor int,    
 dev int,
 qa int,
 savedate datetime
)

insert into @Table values (1,100,22,0, 0, '2008-10-29 11:17:59.527')
insert into @Table values (2,100,5,0, 0, '2008-10-29 11:00:37.030')
insert into @Table values (3,101,22,0, 0, '2009-10-29 11:10:27.687')
insert into @Table values (5,101,44,0, 0, '2008-10-31 12:07:52.917')
insert into @Table values (6,101,32,0, 0, '2009-06-30 08:16:12.343')
insert into @Table values (7,101,44,0, 0, '2009-10-30 10:12:11.369')
insert into @Table values (8,101,44,1, 0, '2009-10-31 10:12:11.369')

;WITH    rows AS
        (
        SELECT  *, ROW_NUMBER() OVER (PARTITION BY ticketid ORDER BY savedate DESC) AS rn
        FROM    @Table
        )
SELECT  rl.sponsor, ro.savedate
FROM    rows rl
CROSS APPLY
        (
        SELECT  TOP 1 rc.savedate
        FROM    rows rc
        JOIN    rows rn
        ON      rn.ticketid = rc.ticketid
                AND rn.rn = rc.rn + 1
                AND rn.sponsor <> rc.sponsor
        WHERE   rc.ticketid = rl.ticketid
        ORDER BY
                rc.rn
        ) ro
WHERE   rl.rn = 1

我在测试数据中为赞助商44 添加了一条记录。

查询将返回行 1 和 7，因为赞助商在行 8 中没有更改。