Message-ID: <433BBC07.5040208@archonet.com>
Date: Thu, 29 Sep 2005 11:03:51 +0100
From: Richard Huxton <dev@archonet.com>
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To: Axel Rau <Axel.Rau@Chaos1.DE>
Cc: pgsql-sql@postgresql.org
Subject: Re: Selecting count of details along with details columns
References: <8365994213849c93946e61ded7d12318@Chaos1.DE>
	<433BA637.7040106@archonet.com>
	<a79cd5e3132c2d4c3430a34e64d93ebe@Chaos1.DE>
In-Reply-To: <a79cd5e3132c2d4c3430a34e64d93ebe@Chaos1.DE>
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Axel Rau wrote:
> 
> Am 29.09.2005 um 10:30 schrieb Richard Huxton:
> 
>> Axel Rau wrote:
>>
>>> SELECT T2.T2_name, COUNT(T1.id) AS xx
>>>     FROM T2, T1
>>>     WHERE T2.id = T1.fk_t2
>>>     GROUP BY T2.T2_name
>>>     HAVING COUNT(T1.id) > 1
>>>     ORDER BY xx DESC;
>>>  t2_name | xx
>>> ---------+----
>>>  T2-N2   |  3
>>>  T2-N3   |  2
>>> (2 rows)
>>> Adding column t1_name to the result set breaks COUNT(T1.id):
>>> SELECT T2.T2_name, T1.T1_name, COUNT(T1.id) AS xx
>>>     FROM T2, T1
>>>     WHERE T2.id = T1.fk_t2
>>>     GROUP BY T2.T2_name, T1.T1_name
>>>     HAVING COUNT(T1.id) > 1
>>>     ORDER BY xx DESC;
>>>  t2_name | t1_name | xx
>>> ---------+---------+----
>>> (0 rows)
>>> How can I do this with pg ?
>>
>>
>> Do what? You don't say what results you are expecting.
>>
>> Do you want:
>> 1. ALL values of T1_name (in which case what count do you want)?
>> 2. The FIRST value of T1_name (in which case what do you mean by first)?
> 
> 
> #1.:
> 
>  t2_name | t1_name | count
> ---------+---------+-------
>  T2-N2   | T1-CCC  |     3
>  T2-N3   | T1-FFF  |     2
>  T2-N2   | T1-BBB  |     3
>  T2-N2   | T1-DDD  |     3
>  T2-N3   | T1-EEE  |     2

Ah - this is two questions:
1. What are the unique (t2_name,t1_name) pairings?
2. How many different (t1.id) values are there for each t2.

So - something like:

SELECT names.T2_name, names.T1_name, counts.num_t2
FROM
(
SELECT DISTINCT T2.T2_name, T1.T1_name
FROM T2,T1
WHERE T2.id = T1.fk_t2
) AS names,
(
SELECT T2.T2_name, COUNT(T1.id) AS num_t2
FROM T2, T1
WHERE T2.id = T1.fk_t2
GROUP BY T2.T2_name
HAVING COUNT(T1.id) > 1
) AS counts
WHERE
   names.T2_name = counts.T2_name
;

You could write the "names" sub-query with a GROUP BY if you wanted of 
course.
-- 
   Richard Huxton
   Archonet Ltd