Re: COUNT Function

From: Carlos <miotromailcarlos_at_netscape.net>
Date: Mon, 17 May 2010 08:30:10 -0700 (PDT)
Message-ID: <f67dfb15-88e5-4245-8701-ff8a6a5b1a14_at_w3g2000vbd.googlegroups.com>

On May 17, 5:01 pm, The Magnet <a..._at_unsu.com> wrote:
> Hi,
>
> I'm tryng to use the Analytical function COUNT(*):
>
> SELECT tag_id, tag_name, tag_count
> FROM (SELECT t.tag_id, tag_name,
>              COUNT(*) OVER (PARTITION BY t.tag_id) tag_count,
>              ROW_NUMBER() OVER (PARTITION BY t.tag_id ORDER BY
> t.tag_id) rnum
>       FROM commentary.article_tags a, commentary.tags t
>       WHERE t.tag_id = a.tag_id(+))
> WHERE rnum = 1;
>
> Problem I am having is that for records in the TAGS table which do not
> match in the ARTICLE_TAGS table are still being returned with 1 row.
> I'm looking for it to return 0, as there were no matches.
>
> Thought it was the way the join was working, but I do not think so as
> I've tried different combos.
>
> Any ideas?

I think i don't get it.

Why not:

CARLOS_at_XE.bequeath> select * from tags;

    TAG_ID TAG_NAME
---------- ----------

         1 TAG 1
         2 TAG 2
         3 TAG 3

CARLOS_at_XE.bequeath> select * from article_tags;

ARTICLE_ID ARTICLE_NAME TAG_ID
---------- ------------ ----------

         1 ARTICLE 11            1
         1 ARTICLE 11            1
         1 ARTICLE 12            1
         2 ARTICLE 21            2

CARLOS_at_XE.bequeath> SELECT t.tag_id,
  2         t.tag_name,
  3         COUNT(a.tag_id) tag_count
  4    FROM article_tags a,
  5         tags t

  6 WHERE t.tag_id = a.tag_id(+)
  7 group by t.tag_id, t.tag_name;

    TAG_ID TAG_NAME TAG_COUNT
---------- ---------- ----------

         3 TAG 3               0
         1 TAG 1               3
         2 TAG 2               1

HTH. Cheers.

Carlos. Received on Mon May 17 2010 - 10:30:10 CDT