Re: MySQL error message...

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mysql knows.  this means your query failed, $command resulted in an invalid
result set.  try adding: or die(mysql_error());  to any call to mysql_query
()....so in your code use this:

$result  = mysql_query("$command",$mysqlHandle) or die(mysql_error());
- and -
$resultw  = mysql_query("$commandw",$mysqlHandle) or die(mysql_error());

that will give you all the info you need.

hth
jeff



                                                                                                                                        
                      Keith Spiller                                                                                                     
                      <larentium@hosthi        To:       php-db@lists.php.net                                                           
                      ve.com>                  cc:       Keith Spiller <larentium@hosthive.com>                                         
                                               Subject:   MySQL error message...                                                
                      06/30/2003 12:47                                                                                                  
                      PM                                                                                                                
                                                                                                                                        
                                                                                                                                        




Hello,

Does anyone know what would cause this message?

Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result
resource in c:\program files\easyphp\www\bod-rse\bod\menu.php3 on line
49...

Here is the code:

<?
 $command = "SELECT * FROM central_fakepaths WHERE menu != '0' ORDER BY
 menu
ASC";
 $result  = mysql_query("$command",$mysqlHandle);
 $path_directory=explode("/", $HTTP_SERVER_VARS["REQUEST_URI"]);

 while ($myrow = mysql_fetch_row($result))
 {
   $count++;

   If   ($path_directory[1]=="$devsite")  { $path = $myrow[9];  $target =
$myrow[10]; }
   else                                   { $path = $myrow[7];  $target =
$myrow[3];  }

   If ($myrow[5] != $count)
   { echo "<img src=\"images/menu-line.gif\" width=\"139\"
height=\"7\"><br>\n";
     $count++;
   }

          // IF MENU VALUE == COMMITTEES
   if ($myrow[2] == "Committees")
   {

       echo "<font face=Verdana size=1><img src='images/menu-line.gif'
width=139 height=7><BR>
             &nbsp; <a href='.' target='_top'>Committees</a><BR>
             <img src='images/menu-line.gif' width=139 height=7><BR>";

  $currdate = date("Y-m",time());
  $cdate    = explode("-" , $currdate);
  $cyear    = $cdate[0];
  $cmonth   = $cdate[1];

  if ($cmonth >=7 AND $cmonth <=12) $fyear  = $cyear + 1;
  else                              $fyear  = $cyear;
                                    $fyear2 = $fyear - 1;
        $syear  = "$fyear2 - $fyear";

    $commandw = "SELECT menuname, type, active, name FROM bod_board_boards
WHERE type='committee' AND active='1' ORDER BY name ASC";
    $resultw  = mysql_query("$commandw",$mysqlHandle);
    while ($myroww = mysql_fetch_row($resultw))
     {
       if ($myroww[0] !="")
        {
   echo "&nbsp; <a href='committees.php?bord=$bord&comm=$myroww[3]'
target='_top'>$myroww[0]</a><BR>";
        }
     }

       //echo "&nbsp; <a href='?bord=$bord' target='_top'>What They
   Do</a>";
   }

   else
   {
     echo "&nbsp; <a href='$path' target='$target'>$myrow[2]</a><br>\n";
   }
 }

?>


Any help would be unendingly appreciated...


Keith


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