Re: [PHP-DB] problem....

[Cole S. Ashcraft]

The error he is getting means that his MySQL resource (in this case the
results of the query) does not exist. Try
$pic1=mysql_fetch_array($pic);
Also, can you give the exact text of the error (line number and all) and line numbers for the entire passage? 

Cole
Shahmat Bin Dahlan wrote:
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" <[EMAIL PROTECTED]>
Date: Wednesday, June 23, 2004 8:58 am
Subject: Re: [PHP-DB] problem....
 

I checked em all they were right
"Daniel Clark" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
   

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" <[EMAIL PROTECTED]> wrote in message
       

news:[EMAIL PROTECTED]>
   

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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