选择每个联系人的最早日期

Question

我有以下查询，它应该只为每个联系人拉最早的日期，但它拉所有可用的日期。 我在这里检查了多个线程，但无法解决。 这正在使用SQL Server 2005。

SELECT DISTINCT o.SubscriberKey, MIN(o.EventDate) as OpenDate
FROM _Open o 
INNER JOIN _Job j
    ON o.JobID = j.JobID
GROUP BY o.SubscriberKey, o.EventDate

目前，我得到这样的结果：

Subscriber 1  17 July 2019 06:04
Subscriber 1  17 July 2019 06:05
Subscriber 1  18 July 2019 04:29
Subscriber 2  18 July 2019 07:04
Subscriber 2  18 July 2019 07:21
Subscriber 2  24 July 2019 05:40

我想实现的目标是：

Subscriber 1  17 July 2019 06:04
Subscriber 2  18 July 2019 07:04

Answer 1

SELECT 
     --you dont want to have a DISTINCT here, you are doing a GROUP BY so its not needed 
    --DISTINCT
    o.SubscriberKey
    , MIN(o.EventDate) as OpenDate
FROM _Open o 
    --Join is not referenced. Thanks @Gordon Linoff
    --INNER JOIN _Job j
    --    ON o.JobID = j.JobID
GROUP BY
    o.SubscriberKey
    --dont group on Event date, this is stopping the MIN function from aggregating the rows.
    --, o.EventDate

Answer 2

您不需要JOIN 。 您需要修复GROUP BY ：

SELECT o.SubscriberKey, MIN(o.EventDate) as OpenDate
FROM _Open o 
GROUP BY o.SubscriberKey;

如果要过滤结果，则只需要JOIN ，但我怀疑情况确实如此。

Answer 3

您只需EventDate group by SubscriberKey并在EventDate聚合。
另外，您正在加入_Job但是您不使用它。
除非您只想获取表中匹配行的最小EventDate ，否则不需要此联接：

SELECT SubscriberKey, MIN(EventDate) as OpenDate
FROM _Open 
GROUP BY SubscriberKey