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SQL query question

От
"Uwe C. Schroeder"
Дата:
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Hash: SHA1


Maybe it's to late for me to think correctly (actually I'm sure of that). I'm
going to ask anyways.
I have a table like

id int4
user_id int4
photo varchar
image_type char(1)

where image_type is either G or X
What I want to do is have ONE query that gives me the count of images of each
type per user_id.
So if user 3 has 5 photos of type G and 3 photos of type X
I basically want to have a result 5,3
It got to be possible to get a query like that, but somehow it eludes me
tonight.

Any pointers are greatly appreciated.

    UC

- --
Open Source Solutions 4U, LLC    2570 Fleetwood Drive
Phone:  +1 650 872 2425        San Bruno, CA 94066
Cell:   +1 650 302 2405        United States
Fax:    +1 650 872 2417
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Re: SQL query question

От
Jonel Rienton
Дата:
Hi Uwe,

I did a solution for you using PLPgSQL,

create or replace function countem() returns varchar as $$
declare
    gcount integer;
    xcount integer;
    result varchar;
begin
         select count(*) into gcount
         from pix where image_type = 'G';

         select count(*) into xcount
         from pix where image_type = 'X';

         select gcount || ', ' || xcount
         into result;

         return result;

end;
$$ LANGUAGE plpgsql;

hope this helps, it's simple and always, there's another (better)
solution
it's my first stab at plpgsql so please bear with me.

-----
Jonel Rienton
http://blogs.road14.com
Software Developer, *nix Advocate

On Feb 3, 2005, at 1:32 AM, Uwe C. Schroeder wrote:

> -----BEGIN PGP SIGNED MESSAGE-----
> Hash: SHA1
>
>
> Maybe it's to late for me to think correctly (actually I'm sure of
> that). I'm
> going to ask anyways.
> I have a table like
>
> id int4
> user_id int4
> photo varchar
> image_type char(1)
>
> where image_type is either G or X
> What I want to do is have ONE query that gives me the count of images
> of each
> type per user_id.
> So if user 3 has 5 photos of type G and 3 photos of type X
> I basically want to have a result 5,3
> It got to be possible to get a query like that, but somehow it eludes
> me
> tonight.
>
> Any pointers are greatly appreciated.
>
>     UC
>
> - --
> Open Source Solutions 4U, LLC    2570 Fleetwood Drive
> Phone:  +1 650 872 2425        San Bruno, CA 94066
> Cell:   +1 650 302 2405        United States
> Fax:    +1 650 872 2417
> -----BEGIN PGP SIGNATURE-----
> Version: GnuPG v1.2.3 (GNU/Linux)
>
> iD8DBQFCAdOMjqGXBvRToM4RApgvAJsEUsdl6hrVGqRwJ+NI7JrqQqQ5GgCgkTQN
> pavTkx47QUb9nr7XO/r/v5k=
> =B3DH
> -----END PGP SIGNATURE-----
>
>
> ---------------------------(end of
> broadcast)---------------------------
> TIP 2: you can get off all lists at once with the unregister command
>     (send "unregister YourEmailAddressHere" to
> majordomo@postgresql.org)
>


Re: SQL query question

От
Jonel Rienton
Дата:
you're right it's late, i better to get to bed myself, i forgot to
throw in the parameter for the user_id in there, i'm sure you can
figure that one out.

regards,

-----
Jonel Rienton
http://blogs.road14.com
Software Developer, *nix Advocate
On Feb 3, 2005, at 1:32 AM, Uwe C. Schroeder wrote:

> -----BEGIN PGP SIGNED MESSAGE-----
> Hash: SHA1
>
>
> Maybe it's to late for me to think correctly (actually I'm sure of
> that). I'm
> going to ask anyways.
> I have a table like
>
> id int4
> user_id int4
> photo varchar
> image_type char(1)
>
> where image_type is either G or X
> What I want to do is have ONE query that gives me the count of images
> of each
> type per user_id.
> So if user 3 has 5 photos of type G and 3 photos of type X
> I basically want to have a result 5,3
> It got to be possible to get a query like that, but somehow it eludes
> me
> tonight.
>
> Any pointers are greatly appreciated.
>
>     UC
>
> - --
> Open Source Solutions 4U, LLC    2570 Fleetwood Drive
> Phone:  +1 650 872 2425        San Bruno, CA 94066
> Cell:   +1 650 302 2405        United States
> Fax:    +1 650 872 2417
> -----BEGIN PGP SIGNATURE-----
> Version: GnuPG v1.2.3 (GNU/Linux)
>
> iD8DBQFCAdOMjqGXBvRToM4RApgvAJsEUsdl6hrVGqRwJ+NI7JrqQqQ5GgCgkTQN
> pavTkx47QUb9nr7XO/r/v5k=
> =B3DH
> -----END PGP SIGNATURE-----
>
>
> ---------------------------(end of
> broadcast)---------------------------
> TIP 2: you can get off all lists at once with the unregister command
>     (send "unregister YourEmailAddressHere" to
> majordomo@postgresql.org)
>


Re: SQL query question

От
Roman Neuhauser
Дата:
# uwe@oss4u.com / 2005-02-02 23:32:28 -0800:
> -----BEGIN PGP SIGNED MESSAGE-----
> Hash: SHA1
>
>
> Maybe it's to late for me to think correctly (actually I'm sure of
> that). I'm going to ask anyways.  I have a table like
>
> id int4
> user_id int4
> photo varchar
> image_type char(1)
>
> where image_type is either G or X
> What I want to do is have ONE query that gives me the count of images
> of each type per user_id.
> So if user 3 has 5 photos of type G and 3 photos of type X
> I basically want to have a result 5,3

    SELECT COUNT(*) FROM t GROUP BY t.user_id, t.image_type

--
If you cc me or remove the list(s) completely I'll most likely ignore
your message.    see http://www.eyrie.org./~eagle/faqs/questions.html

Re: SQL query question

От
Markus Schulz
Дата:
Am Donnerstag, 3. Februar 2005 08:32 schrieb Uwe C. Schroeder:
> Maybe it's to late for me to think correctly (actually I'm sure of
> that). I'm going to ask anyways.
> I have a table like
>
> id int4
> user_id int4
> photo varchar
> image_type char(1)
>
> where image_type is either G or X
> What I want to do is have ONE query that gives me the count of images
> of each type per user_id.
> So if user 3 has 5 photos of type G and 3 photos of type X
> I basically want to have a result 5,3
> It got to be possible to get a query like that, but somehow it eludes
> me tonight.
>
> Any pointers are greatly appreciated.
>
>  UC

select user_id,image_type, count(id)
from <table>
group by user_id,image_type

should do it.

--
Markus Schulz