php - Can only see friends lists if logged in with user id of 1 -

i'm implementing friends list system on site, , i've gotten display friends names have accepted(therefore accepted has value of 1), , when visit dummy accounts can see friends, when logged in dummy account can see either, manually changed user_id in database "1" , logged out , in , discovered it's working user_id of 1, here code. it's not mysqli yet, that's next step.

<h1>friends</h1>  <?php $user_id = user_id_from_username($username); if($_session['user_id'] == $user_id){     $logged_user_id = $_session['user_id'];      $result = mysql_query("select * `friends` `friend_id`='{$logged_user_id}' , `user_id`!='{$logged_user_id}' , `accepted`='1'");     while ($row = mysql_fetch_array($result)){           $friend_id = $row['user_id'];          /*get friend details*/         $fetch_details = mysql_fetch_object(mysql_query("select * `users` `user_id`='{$friend_id}'"));          echo $fetch_details->username;         echo '<br/>';     } }  else if($_session['user_id'] != $user_id){     $user_id = user_id_from_username($username);     $logged_user_id = $user_id;      $result = mysql_query("select * `friends` `user_id`='{$logged_user_id}' , `friend_id`!='{$logged_user_id}' , `accepted`='1'");     while ($row = mysql_fetch_array($result)){           $friend_id = $row['friend_id'];          /*get friend details*/         $fetch_details = mysql_fetch_object(mysql_query("select * `users` `user_id`='{$friend_id}'"));          echo $fetch_details->username;         echo '<br/>';     } } ?> 

tables

friends id(ai) user_id friends_id datemade accepted(enum 0, 1)  users user_id(ai) username profile active 

in first sql query use same variable twice filter:

`friend_id`='{$logged_user_id}' , `user_id`!='{$logged_user_id}' 

this not give expected results.

you add print_r($row) check if expected results database make sure queries want them to.