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php warning: mysql_free_result(): supplied argument is not a valid MySQL result

Posted on 2007-12-01
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2,392 Views
Last Modified: 2013-12-12
I keep getting this warning in a phpbb2 script after adding a mod (i must say that this file hasn't been edited)

Warning: mysql_free_result(): supplied argument is not a valid MySQL result resource in /mounted-storage/home43a/sub004/sc31952-BMDX/www/forum/db/mysql4.php on line 319

This is the code in that line of the file:

            if ( $query_id )
            {
                  unset($this->row[$query_id]);
                  unset($this->rowset[$query_id]);

                  mysql_free_result($query_id); //line 319

                  return true;
            }
            else
            {
                  return false;
            }
      }

How can i fix this??

When i print the content of the variable $query_id I get  this "Resource id #351"
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Comment
Question by:axtur
  • 2
  • 2
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6 Comments
 
LVL 21

Expert Comment

by:nizsmo
ID: 20390120
try and track down where $query_id is declared, and see if it is a connection link to mysql database.

You can temporarily suppress the error message:
@mysql_free_result($query_id);

but off course, you should fix up the error if possible.
0
 

Author Comment

by:axtur
ID: 20390137
It is in this function:


      function sql_freeresult($query_id = 0)
      {
            if( !$query_id )
            {
                  $query_id = $this->query_result;
            }

            if ( $query_id )
            {
                  unset($this->row[$query_id]);
                  unset($this->rowset[$query_id]);
                  mysql_free_result($query_id);

                  return true;
            }
            else
            {
                  return false;
            }
      }
0
 
LVL 21

Expert Comment

by:nizsmo
ID: 20390155
This may be relevant to your problem:
http://www.opentools.de/archive/index.php?t=5330

PLease try and see if this helps.
0
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The purpose of this paper is to provide you background on SQL Server. It’s your self-study guide for learning fundamentals. It includes both the history of SQL and its technical basics. Concepts and definitions will form the solid foundation of your future DBA expertise.

 
LVL 20

Accepted Solution

by:
steelseth12 earned 500 total points
ID: 20390341
try this
function sql_freeresult($query_id = 0)
      {
            if( !$query_id )
            {
                  $query_id = $this->query_result;
            }
 
            if ( $query_id )
            {
                  unset($this->row[$query_id]);
                  unset($this->rowset[$query_id]);
                 
				 if(is_resource($query_id)) {
				 
				  mysql_free_result($query_id);
				
				}
                  return true;
            }
            else
            {
                  return false;
            }
      }

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0
 
LVL 20

Expert Comment

by:steelseth12
ID: 20390376
the @mysql_free_result( suggested by nizsmo only hides the error .
The above solves the problem .
0
 

Author Closing Comment

by:axtur
ID: 31412156
Solved, great answer, it also helped me solve another problem.
Thank you!!!
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