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    <p>2 things:</p> <pre><code>if ($row['username_id'] == 1) </code></pre> <p>you should put that in your sql:</p> <pre><code>$result = mysql_query("SELECT username, friendname FROM tbusers INNER JOIN tbfriends ON tbusers.id = tbfriends.username_id where tbusers.id = ".$yourVariable); </code></pre> <p>In your query, user and friend are linked, and can only be equal.</p> <p>Now, this:</p> <pre><code>$count = $row['friendname'];//friendname is in friends if ($row['id'] == $count)//id is in users </code></pre> <p>is equal to </p> <pre><code>if ( $row['id'] == $row['friendname'] ) </code></pre> <p>this sounds plain wrong. You compare a numerical id with a name. Moreover, your sql query already retrives all friends from users. In the version I showed here, it retrieves only friends of the user you are interested in.</p> <p>finally, you print (echo) the name of the user, not of the friend. In my opinion the following code will do what you want:</p> <pre><code>$result = mysql_query("SELECT friendname FROM tbfriends WHERE username_id = ".$yourUserVariable); while($row = mysql_fetch_array($result)) { echo $row['friendname']; // or better: echo $row['friendname'], '&lt;br&gt;'; } </code></pre> <p>edit: after comment...</p> <p>so <code>if ( $row['id'] == $row['friendname'] )</code> means the user is his own friend. can that happen?</p> <p>this code shall print what you want, friend names.</p> <pre><code>/* $result = mysql_query(" SELECT username as friend_name, friendname as friend_id, username_id as user_id FROM tbusers INNER JOIN tbfriends ON tbusers.id = tbfriends.username_id WHERE tbusers.id = ".$yourVariable); </code></pre> <p>*/</p> <pre><code>$result = mysql_query(" SELECT username as friend_name, friendname as friend_id, username_id as user_id FROM tbusers INNER JOIN tbfriends ON tbusers.id = tbfriends.friendname WHERE tbfriends.username_id = ".$yourVariable); while($row = mysql_fetch_array($result)) { echo $row['friend_name']; // or better: echo $row['friend_name'], '&lt;br&gt;'; } </code></pre>
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    1. POGet a number in one table and print the number's referenced name in another table
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    1. COSorry I forgot to mention friendname is also an id. The person's info is in the users table. The friends table is used to check which persons are friends, so your solution work but I only get the ids of the friens that's why I had if ( $row['id'] == $row['friendname'] )
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    2. COmy head hurts ^^'! you should rename friendname into friend_id, for sake of readability? As well. tbfriends.username_id should be renamed to user_id.
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    3. COIt still doesn't work. Some name are still duplicate and some name that should appear first doesn't.
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