选择每个联系人的最早日期

Question

I have the following query, it should pull just the earliest date for each contact, but it's pulling all available dates. I have reviewed multiple threads here but wasn't able to solve it. This is using 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

Currently, I am getting results like this:

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

And what I would like to achieve:

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

You don't need a JOIN for this. And you need to fix the GROUP BY :

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

You only need the JOIN if it is filtering results, but I doubt that is the case.

Answer 3

You only need to group by SubscriberKey and aggregate on EventDate .
Also you are joining _Job but you don't use it.
Unless you only want to get the minimum EventDate of the matching rows of the tables there is no need for this join:

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