yet another argument for sub-selects :)
Paul DuBois wrote:
>
> At 2:31 PM -0800 3/22/01, Daren Cotter wrote:
> >I have a table, which keeps track of member information (including which
> >member referred the member). To get a count of the # of referrals for member
> >25, my query is:
> >
> >SELECT COUNT(*) FROM members WHERE ref1 = 25;
> >
> >To get a list of the top referers and the # of referrals they have, my query
> >is:
> >
> >SELECT DISTINCT(ref1) AS member_id, COUNT(*) AS count FROM members GROUP BY
> >ref1 ORDER BY count DESC LIMIT 100
> >
> >However, what I need, is a list of the top referers, along with their member
> >information...name, email, password, etc. I tried using the following query,
> >as I read about it in the MySQL manual, but it doesn't work:
>
> You're asking for a summary as well as a listing. You'll need two queries.
> One to get the member_id for the top 25 referers, another to pull the
> information for those members. You can use ...WHERE member_id IN (list)
> on the second query, where "list" is a comma separated list of the member_id
> values. Probably best to put this stuff in a Perl script or something so
> that you can manipulate the list and construct the queries easily.
>
> >
> >SELECT DISTINCT(a.ref1) AS member_id, count(*) AS count, b.password,
> >concat(UCASE(SUBSTRING(b.first_name,1,1)),
> >LCASE(SUBSTRING(b.first_name,2,LENGTH(b.first_name)))) AS name, b.email,
> >b.html_mail, b.ref1, DATE_FORMAT(b.signup_date, '%b %e, %Y') AS signup_date
> >FROM members AS a, members AS b WHERE a.active_member = 'Y' AND a.ref1 =
> >b.member_id GROUP BY a.ref1 ORDER BY count DESC LIMIT 10
> >
> >This gives me correct info for the distinct a.ref1 and count fields, and
> >produces data for the rest of the fields, but it is not actually that
> >member's data. Is this possible to do with one query? If I want to get the
> >top 100 referers' data, I don't want to do 100 separate queries. Please
> >help!
>
> --
> Paul DuBois, [EMAIL PROTECTED]
>
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