Re: problem....

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Yes, your right, but I also think his MySQL SQL statement needs to be
checked.

Maybe can also try to add this right after the mysql_query function call.

echo mysql_errno() . ": " . mysql_error() . "\n"; 


----- Original Message -----
From: "Cole S. Ashcraft" <cashcraft@xxxxxxxxxxxxxxxxxx>
Date: Wednesday, June 23, 2004 9:47 am
Subject: Re:  problem....

> 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/)
> >>To unsubscribe, visit: http://www.php.net/unsub.php
> >>
> >>
> >>    
> >>
> >
> >  
> >
> 
> 
> -- 
> This message has been scanned for viruses and
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> 
> 

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