Re: problem....

From:	Cole S. Ashcraft	Date:	Wed, 23 Jun 2004 01:57:53 +0000
Subject:	Re: problem....
References:	1 2	Groups:	php.db
Request:	Send a blank email to php-db+get-34878@lists.php.net to get a copy of this message

I need the exact error message. Copy Paste it from the browser. I also need line numbers. Cole water_foul wrote:

i fixed those things and it didn't fix it :( :( :( :( :( :( :( :( :( is there a icq chatroom for php?) "Shahmat Bin Dahlan" <shahmatd@sains.com.my> wrote in message news:16780f91a.f91a16780@mail.sarawaknet.gov.my...
Your SQL statement:
'SELECT     Rune, username FROM         RuneRunner
RuneRunner_1 WHERE     (User_ID = 3)'
What is "RuneRunner" and "RuneRunner_1"? Are these two different tables. If it is, you might want to separate them with a comma. Why not try getting rid of the brackets surrounding "User_ID=3"? And also wrap single quotes around the digit "3". ----- Original Message ----- From: "water_foul" <phpdbnews@aaron.aichlmayr.net> Date: Wednesday, June 23, 2004 8:58 am Subject: Re: [PHP-DB] problem....
I checked em all they were right "Daniel Clark" <dclark@nwlink.com> wrote in message news:18323.65.203.232.102.1087951969.squirrel@webmail.pacifier.com...
     
Sounds like it doesn't like your SQL statement. Perhaps a field
       
or table
     
name is incorrect?
       
Warning: mysql_fetch_array(): supplied argument is not a valid
         

MySQL> > result

resource in ....... "Daniel Clark" <dclark@nwlink.com> wrote in message

news:16692.65.203.232.102.1087950853.squirrel@webmail.pacifier.com...>

What error are you getting?

why doesn't this work:

$pic=mysql_query('SELECT     Rune, username FROM
             
RuneRunner> >> > RuneRunner_1 WHERE     (User_ID = 3)',$connection);

$pic=mysql_fetch_array($pic);

             
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