I need help with this query. The databases looks like this:
FriendsUsers<?php
//suggested friends
$suggestedHTML = '';
$sql = "SELECT username, avatar FROM users LEFT JOIN friends ON
users.id!=friends.user2_id";
$query = mysqli_query($db_conx, $sql);
while($row = mysqli_fetch_array($query, MYSQLI_ASSOC)) {
$suggested_username = $row["username"];
$suggested_avatar = $row["avatar"];
if($suggested_avatar != ""){
$suggested_pic = 'users/'.$suggested_username.'/'.$suggested_avatar.'';
} else {
$suggested_pic = 'images/avatardefault.jpg';
}
$suggestedHTML .= '<div class="friend-div"><a href="user.php?u='.$suggested_username.'"><img class="friendpics" src="'.$suggested_pic.'" alt="'.$suggested_username.'" title="'.$suggested_username.'"><span id="friend-user">'.$suggested_username.'<span></a></div>';
}
?>
I need the username and avatar to output on the screen where friend.accepted is != 1, but the query just runs through every record in the friends table.
What is the right approach?
I only want to output the username and avatar for the records in users where there is no friend relationship.