mysql: showall() = select * from sometable; class.php: class EMP_LOGIN { function EMP_LOGIN() { } function showall() { $myrset = mysql_query("call showall()"); return $myrset; } function getemp($somevar) { somecommands; } } page1.php: $myobj = new EMP_LOGIN(); $myrset = $myobj->showall(); $myrow = mysql_fetch_array($myrset); echo($myrow['EMP_ID']); page2.php: //First procedure to get details for a single employee $myobj = new EMP_LOGIN(); $myrset = $myobj->getemp($_SESSION['EMP_ID']); $myrow = mysql_fetch_array($myrset); echo "Welcome! ".$myrow['FIRSTNAME']; echo " ".$myrow['LASTNAME']; //This procedure is to get dashboard with all employee details $myrset = $myobj->showall(); $myrow = mysql_fetch_array($myrset); echo($myrow['EMP_ID']); Page1.php works well but page2.php returns Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource What went wrong. Is it wrong to access two objects in a single page. Help needed.