Re: SUM() & GROUP BY

Thanks Huxton,

Sorry for not explaining fully here is what I would like to achieve:

When I do:

SELECT s.pd_geo, COUNT(s.pd_geo) FROM ser s GROUP BY s.pd_geo;
OUTPUT:
pd_geo | count
----------------------
1 | 49
2 | 39
3 | 6
4 | 54

It outputs the number of entries the Divisions have made

Here is what Division table contains:

SELECT * FROM ser_divisions;

divisions_name | divisions_id
---------------------------------------
DEE | 3131
DEPI | 3133
DED | 3134
GBH | 3136

Now I would like to get to know how each Division answered i.e.

SELECT s.pd_geo, COUNT(s.pd_geo)
FROM ser s
WHERE s.ser_divisions = '3131'
GROUP BY s.pd_geo;

output:

pd_geo | count
----------------------
1 | 9
2 | 2
3 | 6
4 | 5

But this is the output I intend to get:

divisions_name | pd_geo | count
-----------------------------------------------
DEE | 1 | 9
DEE | 2 | 2
DEE | 3 | 6
DEE | 4 | 5

How do I achieve the above results please do advice thanks again.

Kind Regards
+-----------------------------------------------------+
| Martin W. Kuria (Mr.) martin(dot)kuria(at)unon(dot)org
+----------------------------------------------------+

>From: Richard Huxton <dev(at)archonet(dot)com>
>To: Martin Kuria <martinkuria(at)hotmail(dot)com>
>CC: olly(at)lfix(dot)co(dot)uk, middink(at)indo(dot)net(dot)id, pgsql-sql(at)postgresql(dot)org
>Subject: Re: [SQL] SUM() & GROUP BY
>Date: Fri, 07 May 2004 09:00:43 +0100
>
>Martin Kuria wrote:
>>Hi again I have two tables I would like to query i.e. service table
>>and division table
>>
>>SELECT s.pd_geo, COUNT(s.pd_geo) FROM ser s GROUP BY s.pd_geo;
>>OUTPUT:
>>pd_geo | count
>>----------------------
>> 1 | 49
>> 2 | 39
>> 3 | 6
>> 4 | 54
>>
>>SELECT d.divisions_name, d.divisions_id)
>>FROM ser s, ser_divisions d
>>WHERE d.divisions_id = s.ser_divisions;
>>
>>division_name | divisions_id
>>--------------------------------------
>> DEC | 6
>> DEPI | 7
>> DRC | 8
>>
>>How can I create a query that displays How the divisions answered
>>the question please do assist.
>
>Martin - you'll need to explain exactly what you want. Can you show
>what outputs you would like given the above data?
>
>--
> Richard Huxton
> Archonet Ltd

_________________________________________________________________
Tired of spam? Get advanced junk mail protection with MSN 8.
http://join.msn.com/?page=features/junkmail