Proble with an if/else

[graham23s]

Hi Guys,

 

i htought i had this query pretty much down , but it's not working, basically i want to check if a logged in user has alreaady added someone as a friend or not if so "you have already added this user as a friend" else enter the users info into mysql

 

code:

 

<?php 
     // add to friends...////////////////////////////////////////////////////////////////
     $friend_to_adds_id = $_GET['id'];
     
     // the logged in users id...////////////////////////////////////////////////////////
     $query1 = "SELECT `id` FROM `membership` WHERE `username`='$member'";
     $result1 = mysql_query($query1) or die (mysql_error());
     $row = mysql_fetch_array($result1) or die (mysql_error());
     
     // get the logged in users id in a variable...//////////////////////////////////////
     $logged_in_users_id = $row['id'];
     
     // see if the friend has already been added...//////////////////////////////////////
     $query0 = "SELECT `friend_id` FROM `friends` WHERE `friend_id`='$friend_to_adds_id'";
     $result0 = mysql_query($query0) or die (mysql_error());
     
     if ($result0 == 0) {
     
     // all good? lets insert it into the database...////////////////////////////////////
     $query2 = "INSERT INTO `friends` (`user_id`,`friend_id`) VALUES ('$friend_to_adds_id','$logged_in_users_id')";
     $result2 = mysql_query($query2) or die (mysql_error()); 
     
     if ($result2) {
     
          echo "<br /><b>Friend Added Successfully! You Can Remove Them Via The My Account Page.</b><br /><br />";
     
     }
          
     } else {
     
          echo "<br/><b>You Have Already Added This User As A Friend!</b><br /><br />"; 
     
     } ## end else...////////////////////////////////////////////////////////////////////
?>

 

once i delete the entries from mysql (for testing) it always says "You Have Already Added This User As A Friend!"

 

any help would be great

 

cheers

 

Graham

 

[chigley]

<?php 
     // add to friends...////////////////////////////////////////////////////////////////
     $friend_to_adds_id = $_GET['id'];
     
     // the logged in users id...////////////////////////////////////////////////////////
     $query1 = "SELECT `id` FROM `membership` WHERE `username`='$member'";
     $result1 = mysql_query($query1) or die (mysql_error());
     $row = mysql_fetch_array($result1) or die (mysql_error());
     
     // get the logged in users id in a variable...//////////////////////////////////////
     $logged_in_users_id = $row['id'];
     
     // see if the friend has already been added...//////////////////////////////////////
     $query0 = "SELECT `friend_id` FROM `friends` WHERE `friend_id`='$friend_to_adds_id'";
     $result0 = mysql_query($query0) or die (mysql_error());
     
     if (!$result0) {
     
     // all good? lets insert it into the database...////////////////////////////////////
     $query2 = "INSERT INTO `friends` (`user_id`,`friend_id`) VALUES ('$friend_to_adds_id','$logged_in_users_id')";
     $result2 = mysql_query($query2) or die (mysql_error()); 
     
     if ($result2) {
     
          echo "<br /><b>Friend Added Successfully! You Can Remove Them Via The My Account Page.</b><br /><br />";
     
     }
          
     } else {
     
          echo "<br/><b>You Have Already Added This User As A Friend!</b><br /><br />"; 
     
     } ## end else...////////////////////////////////////////////////////////////////////
?>

 

Try that?

[graham23s]

Hi Mate,

 

just tried the code it's still the same (nothing in mysql) but it says "You Have Already Added This User As A Friend!", crazy the code looks pretty straight forward aswell.

 

Graham

[cooldude832]

could you possibly be looking for a blank string??

[lee20]

Your first query ($query0) only checks on friend_id. That would mean that everyone can only have 1 friend!

 

 
$query0 = "SELECT `friend_id` FROM `friends` WHERE friend_id`='$friend_to_adds_id'";
// SHOULD BE
$query0 = "SELECT `friend_id` FROM `friends`
                WHERE friend_id`='$friend_to_adds_id' AND user_id = '$logged_in_users_id'";
// BUT REALLY MAY NEED TO BE
$query0 = "SELECT `friend_id` FROM `friends`
                WHERE (friend_id`='$friend_to_adds_id' AND user_id = '$logged_in_users_id')
                OR (friend_id`='$logged_in_users_id' AND user_id = '$friend_to_adds_id')";

 

If you used the middle query above, then the order of your INSERT appears to be backwards as well:

$query2 = "INSERT INTO `friends` (`user_id`,`friend_id`) VALUES ('$friend_to_adds_id','$logged_in_users_id')";
// SHOULD BE
$query2 = "INSERT INTO `friends` (`user_id`,`friend_id`) VALUES ('$logged_in_users_id','$friend_to_adds_id')";

 

You could bypass the whole issue by creating a unique index on user_id and friend_id, and use the following:

 

// the friend with the lowest id will be stored in user_id, and the other stored in friend_id
// ie: user_id = 1, friend_id = 2
$friends = array($logged_in_users_id, $friends_to_adds_id);
asort($friends);
$query2 = "REPLACE INTO `friends` (`user_id`, `friend_id`) VALUES ('{$friends[0]}', '{$friends[1]}'";

 

Cheers