mysql error! this is pissin me off!

[boo_lolly]

error!

[b]Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/httpd/vhosts/hiphopjam4kidz.com/www/merchandise/includes/item_display.inc on line 27[/b]

what does that mean?!?

here's lines 20 - 35 of item_display.inc:

[code=php:0]
[code]
          </font></b>
          </td>
          <?
          // Now lets see if we need to start a new row
          $tot = 0;
          $rtot = 0;
          $rows = $config[rows_display];
          $cols = $config[cols_display];
          while ($i = mysql_fetch_array($iQuery)) {      //<---- LINE 27
            if(($tot % $config[cols_display]) == 0) {
            print "        </tr><tr>";
              }
          // New row accomplished if needed
            print "          <td align=center class=\"header\">\n";
            print "          <table border=0 width=\"100%\">";
            print "          <tr>";
print "          <td class=\"header\" align=left width=\"200\">";
[/code]
[/code]

[chiprivers]

Might be a stupied question but have you declared the variable $iQuery and if so is it valid?

[fenway]

Yes, having the contents of this query would be necessary to answer the question, though MySQL is already telling you it's invalid.

[arianhojat]

i am gonna guess $iQuery is a string and not a mysql_result($query).

also should this be $config['cols_display'] or $config[$cols_display], otherwise cols_display is a static var i think.

[wildteen88]

If you are getting this error then there is either a problem with your SQL query or the $iQuery variable does not exists or is mistyped.

Could you post the part of the code where you create the $iQuery variable.