This is the code I'm using and I'm a newbie so this is still a lil
challenging for me:
$numberquery = mysql_query("SELECT SUM(s.groundballs) AS grounders FROM
stats s LEFT JOIN rahs r ON s.name = r.name WHERE (s.position =
'$position')");
while($number=mysql_fetch_array($numberquery)){
printf ("Total groundballs for $position: %s</font></td>",$number["grounders"]);
}
Thanks!
--------------------------------------------
-Alex "Big Al" Behrens
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----- Original Message -----
From: "Paul DuBois" <[EMAIL PROTECTED]>
To: "Alex Behrens" <[EMAIL PROTECTED]>; <[EMAIL PROTECTED]>
Sent: Wednesday, March 27, 2002 9:45 PM
Subject: Re: SELECT frmo two tables
> At 20:14 -0600 3/27/02, Alex Behrens wrote:
> >hey guys,
> >
> >this mysql line doesn't seem to work:
> >
> >SELECT SUM(s.groundballs) AS grounders FROM stats s LEFT JOIN rahs r ON
> >s.name = r.name WHERE (s.position = '$position')
> >
> >why not? is there somethign wrong with my syntax? it says this error:
> >"Supplied argument is not a valid MySQL result resource"
>
> You're leaving out details, namely the surrounding PHP code from within
> which you're executing this statement. My guess is that you didn't
> check the result from mysql_query() to see whether or not it succeeeded.
> Do so, and if it fails, print the value of mysql_error() to see what it
> says.
>
> >
> >Thanks!
> >--------------------------------------------
> >-Alex "Big Al" Behrens
> >E-mail: [EMAIL PROTECTED]
> >Urgent E-mail: [EMAIL PROTECTED] (Please be brief!)
> >Phone: 651-482-8779
> >Cell: 651-329-4187
> >Fax: 651-482-1391
> >ICQ: 3969599
> >Owner of the 3D-Unlimited Network:
> >http://www.3d-unlimited.com
> >Send News:
> >[EMAIL PROTECTED]
>
>
>
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