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Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in

Posted on 2011-02-21
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469 Views
Last Modified: 2012-05-11
Cannot figure out why I am getting this error with this code, driving me crazy. Here is even the weirder thing. The code executes correctly, it does exactly what its supposed to do but i get this error. I even add the or die mysql error thing after the query to show me more detail and it shows me nothing. What the heck am I doing wrong?
<? include ('connect.php'); ?>
<? session_start(); ?>
<?
if ($session_user_logged = true) {
	$sql = "SELECT * FROM messages WHERE message_id = '$message_id'";
	$result = mysql_db_query($db,$sql,$cid);
	while ($row = mysql_fetch_array($result)) {
		if (!empty($row['hide'])) {
			$sql = "DELETE FROM messages WHERE message_id = '$message_id'";
			$result = mysql_db_query($db,$sql,$cid);
			if ($result) {
				echo '<meta http-equiv="refresh" content="0;url=../messages.php">';
			} else {
				session_register("error");
				$error = "There was a problem trying to post your status, please try again.";
				echo '<meta http-equiv="refresh" content="0;url=../messages.php">';
			}
		} else {
			$sql = "UPDATE messages SET hide = '$session_user_id' WHERE message_id = '$message_id'";
			$result = mysql_db_query($db,$sql,$cid);
			if ($result) {
				echo '<meta http-equiv="refresh" content="0;url=../messages.php">';
			} else {
				session_register("error");
				$error = "There was a problem trying to post your status, please try again.";
				echo '<meta http-equiv="refresh" content="0;url=../messages.php">';
			}
		}
	}
} else {
	session_register("error");
	$error = "You have to be logged in as the correct user to use this.";
	echo '<meta http-equiv="refresh" content="0;url=../index.php">';
}
?>

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Question by:cbielich
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6 Comments
 
LVL 36

Expert Comment

by:Loganathan Natarajan
ID: 34948943
try like this,
<? include ('connect.php'); ?>
<? session_start(); ?>
<?
if ($session_user_logged = true) {
	$sql = "SELECT * FROM messages WHERE message_id = '$message_id'";
	$result = mysql_db_query($db,$sql,$cid) or die("error in connection ".mysql_error());
	
	$num_rows = mysql_num_rows($result);
	
	if ($num_rows > 0) {
		while ($row = mysql_fetch_array($result)) {
			if (!empty($row['hide'])) {
				$sql = "DELETE FROM messages WHERE message_id = '$message_id'";
				$result = mysql_db_query($db,$sql,$cid);
				if ($result) {
					echo '<meta http-equiv="refresh" content="0;url=../messages.php">';
				} else {
					session_register("error");
					$error = "There was a problem trying to post your status, please try again.";
					echo '<meta http-equiv="refresh" content="0;url=../messages.php">';
				}
			} else {
				$sql = "UPDATE messages SET hide = '$session_user_id' WHERE message_id = '$message_id'";
				$result = mysql_db_query($db,$sql,$cid);
				if ($result) {
					echo '<meta http-equiv="refresh" content="0;url=../messages.php">';
				} else {
					session_register("error");
					$error = "There was a problem trying to post your status, please try again.";
					echo '<meta http-equiv="refresh" content="0;url=../messages.php">';
				}
			}
		}
	} // for num row
} else {
	session_register("error");
	$error = "You have to be logged in as the correct user to use this.";
	echo '<meta http-equiv="refresh" content="0;url=../index.php">';
}
?>

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0
 
LVL 36

Expert Comment

by:Loganathan Natarajan
ID: 34948946
which line the error occur?
0
 
LVL 17

Expert Comment

by:Shinesh Premrajan
ID: 34948954
It seems, the debugging is disabled in your script. In the top of the page add these two lines

error_reporting(E_ALL ^ E_NOTICE);
ini_set("display_errors",1);

This will show up which line is causing the error.

Hope this Helps

0
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LVL 1

Author Comment

by:cbielich
ID: 34948955
Here is what prints out


Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/content/g/u/i/guildfit/html/eaccountabilities/includes/remove_message.php on line 11

That's it
0
 
LVL 17

Accepted Solution

by:
Shinesh Premrajan earned 500 total points
ID: 34948963
The problem is with the $result  variable, You are using the $result variable in the

while ($row = mysql_fetch_array($result)) {

Also you are reassigning it in the delete and update statement, that naturally wont work.


$sql = "DELETE FROM messages WHERE message_id = '$message_id'";
$result_del = mysql_db_query($db,$sql,$cid);
if ($result_del) {

same you need to do for the update statement.

Hope this helps
0
 
LVL 1

Author Comment

by:cbielich
ID: 34948973
OMG dont take this the wrong way but I love you, IT WORKED!!!!!!
0

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