Re: problem....

["Cole S. Ashcraft" <cashcraft@xxxxxxxxxxxxxxxxxx>]

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" <phpdbnews@xxxxxxxxxxxxxxxxxxx>
> Date: Wednesday, June 23, 2004 8:58 am
> Subject: Re:  problem....
>
>  
>
> > I checked em all they were right
> > "Daniel Clark" <dclark@xxxxxxxxxx> wrote in message
> > news:18323.65.203.232.102.1087951969.squirrel@xxxxxxxxxxxxxxxxxxxxxxx
> >    
> >
> > > 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@xxxxxxxxxx> wrote in message
> > > >
> > > >        
> > news:16692.65.203.232.102.1087950853.squirrel@xxxxxxxxxxxxxxxxxxxxxxx>
> >    
> >
> > > 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);
> > > > > >            
> > --
> > PHP Database Mailing List (http://www.php.net/)
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> >
> >
> >    
>
>  



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