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Can somebody help with this query?

 
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revdpoel

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Joined: 01 Nov 2006
Posts: 56

PostPosted: Thu Sep 02, 2010 7:15 pm    Post subject: Can somebody help with this query?
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I have a table which consists of the following fields

automaat-id
automaat-type
automaat-status
contract-id

If for 1 contract-id more than 1 (so 2 or more) different automaat-id occurs with apparaat-status 'A' and automaat-type = 'Pin' or 'Comb' count this as 1 (let's call this X)
I need to have the total of X

What's the best way?
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dbzTHEdinosauer

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Joined: 20 Oct 2006
Posts: 6968
Location: porcelain throne

PostPosted: Thu Sep 02, 2010 8:01 pm    Post subject:
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DB2 tables do not have fields, they have columns.

Also, column names use underscores, not hyphens.

what have you tried. If you say nothing, that is probably the number of responses that you will receive.
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revdpoel

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Joined: 01 Nov 2006
Posts: 56

PostPosted: Thu Sep 02, 2010 10:00 pm    Post subject:
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I don't think this is a rather helpful and kind answer

I haven't tried anything since I don't know where to start
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dick scherrer

Site Director


Joined: 23 Nov 2006
Posts: 19270
Location: Inside the Matrix

PostPosted: Thu Sep 02, 2010 11:41 pm    Post subject:
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Hello,

Quote:
I haven't tried anything since I don't know where to start
Then you need to invest some time (and maybe some $) on SQL training/education/book(s).

This kind of material is part of most basic SQL classes/texts.

We can help when there are problems, but we are not able to "teach" from scratch.
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sqlcode1

Active Member


Joined: 08 Apr 2010
Posts: 578
Location: USA

PostPosted: Fri Sep 03, 2010 1:01 am    Post subject:
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revdpoel,

Try this untested....

Code:
SELECT SUM(CNT) FROM                                           
        (SELECT COUNT(DISTINCT(CONTRACT-ID)) AS CNT
         FROM TABLE                                         
         WHERE APPARAAT-STATUS = 'A'                           
         AND   AUTOMAAT-TYPE IN ('PIN','COMB')                 
         GROUP BY CONTRACT-ID HAVING COUNT(AUTOMAAT-ID) > 1) A


Thanks,
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