problem with code , probly really simple!

[Glenskie]

not working :/

error reads "mysql_error() expects parameter 1 to be resource" on the line that says RIGHT HERE THIS LINE and its commented out for you... whats wrong?

 

 

<?php
if ($_POST["request"] == "requestFriendship") {

    $mem1 = preg_replace('#[^0-9]#i', '', $_POST['mem1']); 
    $mem2 = preg_replace('#[^0-9]#i', '', $_POST['mem2']); 
//
if (!$mem1 || !$mem2 || !$thisWipit) {
	echo  'Error: Missing data';
    	exit(); 
}
//
if ($mem1 == $mem2) {
    	echo  'Error: You cannot add yourself as a friend';
    	exit(); 
} 


//See if already friends
// RIGHT HERE THIS LINE $sql1 = "SELECT id FROM friends WHERE mem2='$mem1' AND mem1='$mem2' LIMIT 1";
$result1 = mysql_query($sql1) or die (mysql_error("sql1 failed")); 
$data1 = mysql_fetch_row($result1);
    if ($data1[0]) {
	echo  'This member is already your Friend'; 
	exit(); 
}
//Check if user already requested to be friends
$sql2 = "SELECT id FROM friends_requests WHERE mem1='$mem1' AND mem2='$mem2' Limit 1";
$result2 = mysql_query($sql2) or die (mysql_error("sql2 failed")); 
$data2 = mysql_fetch_row($result2);
    if ($data2[0]) {
	echo '<img src="images/round_error.png" width="20" height="20" alt="Error" /> You have a Friend request pending for this member. Please be patient.';
        exit(); 
}
//Check if friend already requested to be friends
$sql3 = "SELECT id FROM friends_requests WHERE mem1='$mem2' AND mem2='$mem1' Limit 1";
$result3 = mysql_query($sql3) or die (mysql_error("sql3 failed"));  
$data3 = mysql_fetch_row($result3);
    if ($data3[0]) {
	echo '<img src="images/round_error.png" width="20" height="20" alt="Error" /> This user has requested you as a Friend already! Check your Requests on your profile.';
        exit(); 
}
//If still here make insert	
$mem1=mysql_real_escape_string($mem1);
$mem2=mysql_real_escape_string($mem2);
$sql = mysql_query("INSERT INTO friends (mem1, mem2, timedate) VALUES('$mem1','$mem2',now())") or die (mysql_error("Friend Request Insertion Error"));
echo '<img src="images/round_success.png" width="20" height="20" alt="Success" /> Friend request sent successfully. This member must approve the request.';
    exit(); 
}
?>

[Pikachu2000]

Why do you have the line commented out? Of course you'll get an error when you execute a query with an empty query string . . .

[Pikachu2000]

Oh, wait a minute. it's mysql_error() that's complaining. mysql_error doesn't take a string as an argument, it either takes a connection resource, or no argument and it attempts to use the last connection resource that was used.

[Glenskie]

im just commenting it out so you could find it...

[Glenskie]

so what is the error?

 

[PFMaBiSmAd]

so what is the error?

 

 

Reply #2 states what is wrong with your code.