Re: problem....

Please post the table definition to the mailing list so we can help you more quickly. Also, is there a table called RuneRunner_1 or are you meaning to use RuneRunner_1 as an alias? Odds are, your problem is not PHP related, it's an error in your SQL syntax. Jon 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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