[RESOLVED] mysql_fetch_object

millan

ok so im trying to code something and this is the error i get

Warning: mysql_fetch_object(): supplied argument is not a valid MySQL result resource in /home/realwars/domains/realwars.nl/public_html/crew/helpmenu1.php on line 17
php

and this is the file where the error comes from

<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Admin / Crew page - My Profile</title>
<link href="loginmodule.css" rel="stylesheet" type="text/css" />
</head>
<body>
<p><?php


$dbconn = mysql_connect( "localhost", "user_site", "password");
mysql_select_db("user_site", $dbconn);

$id3 = $_SESSION['SESS_MEMBER_ID'];
$query2 = mysql_query("SELECT isstaff5,isstaff4,isstaff,isstaff2,isstaff3,isstaffleader2 FROM site_user WHERE userID='$id3'");
while ($list2 = mysql_fetch_object($query2)) { 
$isstaff5 = "$list2->isstaff5";
$isstaff4 = "$list2->isstaff4";
$isstaff = "$list2->isstaff";
$isstaff2 = "$list2->isstaff2";
$isstaff3 = "$list2->isstaff3";
$isstaffleader2 = "$list2->isstaffleader2";

echo "<tr>";
echo "<td><br><b><font color=\"red\">Suboptions: </font></b></b></td>";
if ($isstaffleader2 == 1 OR $isstaff5 == 1){
echo "<td><b> | <a href=\"addac1.php\">Add Member</a></b></td>";
echo "<td><b> | <a href=\"delac1.php\">Del Member</a></b></td>";
}
}
?>

Any help?

laserlight

Check the database server connection, the database selection, and that the query is correct and succeeded.

Oh, and why are you using the MySQL extension? You should be using something like the PDO extension or the MySQLi extension.

halojoy

Maybe you forgot session_start();

Note session_start(); must be before any HTML is output.
Before:
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN">
or any other print/echo

<?php

session_start(); // to get the $_SESSION['SESS_MEMBER_ID'] down there


?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Admin / Crew page - My Profile</title>
<link href="loginmodule.css" rel="stylesheet" type="text/css" />
</head>
<body>
<p><?php


$dbconn = mysql_connect( "localhost", "user_site", "password");
mysql_select_db("user_site", $dbconn);

$id3 = $_SESSION['SESS_MEMBER_ID']; 

// more ....
?>

millan

halojoy;10943892 wrote:

Maybe you forgot session_start();

Note session_start(); must be before any HTML is output.
Before:
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN">
or any other print/echo

<?php

session_start(); // to get the $_SESSION['SESS_MEMBER_ID'] down there


?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Admin / Crew page - My Profile</title>
<link href="loginmodule.css" rel="stylesheet" type="text/css" />
</head>
<body>
<p><?php


$dbconn = mysql_connect( "localhost", "user_site", "password");
mysql_select_db("user_site", $dbconn);

$id3 = $_SESSION['SESS_MEMBER_ID']; 

// more ....
?>

No, i did not forget about that it just that i didn't post it here :d

Well i check the query's and post a feedback hope it will work...

halojoy

Good.
Well, you check query.

One thing I should change, as userID most probably is an INTEGER
WHERE userID=$id3

Also such error we should catch using [man]mysql_error/man;
right after the query
but before the fetch (with your error trouble)

$query2 = mysql_query("SELECT isstaff5,isstaff4,isstaff,isstaff2,isstaff3,isstaffleader2 
FROM site_user WHERE userID=$id3");

//will give hint why there was an error and eventual syntax error
if( mysql_error() ) exit( mysql_error() );

while ($list2 = mysql_fetch_object($query2)) {

$isstaff5 = "$list2->isstaff5";

millan

halojoy;10943899 wrote:
Good.
Well, you check query.

One thing I should change, as userID most probably is an INTEGER
WHERE userID=$id3

Also such error we should catch using [man]mysql_error/man;
right after the query
but before the fetch (with your error trouble)
$query2 = mysql_query("SELECT isstaff5,isstaff4,isstaff,isstaff2,isstaff3,isstaffleader2 
FROM site_user WHERE userID=$id3");

//will give hint why there was an error and eventual syntax error
if( mysql_error() ) exit( mysql_error() );

while ($list2 = mysql_fetch_object($query2)) {

$isstaff5 = "$list2->isstaff5";

It's fixed now :p. Thanks alot for help