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Home/ Questions/Q 8598685
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Editorial Team
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Editorial Team
Asked: June 12, 20262026-06-12T01:12:47+00:00 2026-06-12T01:12:47+00:00

When I try to access profile.php?u=destiny //$result = mysql_query(‘SELECT name FROM $imageresult = mysql_query(SELECT

  • 0

When I try to access profile.php?u=destiny

//$result = mysql_query('SELECT name FROM 
$imageresult = mysql_query("SELECT name FROM imagetable WHERE id = '$id'") or die(mysql_error());
$u = mysql_result($imageresult, 0 ,"name") or die(mysql_error());
//error_reporting(E_ALL);
if (isset($id) && (!isset($u))) {
}

Warning: mysql_result() [function.mysql-result]: Unable to jump to row
0 on MySQL result index 5 in profile.php on line 11

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-12T01:12:48+00:00Added an answer on June 12, 2026 at 1:12 am

    This warning means that there is no row in the $imageresult var. Check it out, this should work:

    $imageresult = mysql_query("SELECT name FROM imagetable WHERE id = '$id'") or die(mysql_error());
    if (mysql_num_rows($imageresult) > 0) {
      $u = mysql_result($imageresult, 0 ,"name") or die(mysql_error());
      if (isset($id) && (!isset($u))) {
      }
    }
    
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