hi , in code there Warning

Warning: mysql_fetch_array() expects parameter 1 to be resource, string given in C:\xampp\htdocs\Join.php on line 69

but i can not solving where exactly >>

can you help me where the error in this code .

<?php 

   51. include("connect.php");

   53. $email = mysql_query("select MemberEmail from members where MemberID= '$id' ");
   54. while ($row = mysql_fetch_array($email))
   55.  {   

   57.  $memEmail=$row['MemberEmail'];

   }


  62.  $sql = mysql_query("select * from ninvite where recieverMemberEmail ='$memEmail'  and viwed= '0' order by  RoomID desc");

  64.  $num =mysql_num_rows($sql);


  67.  if ($sql and $num >0 )
      {
  69.   while($row=mysql_fetch_array($sql))
       {
  71.   $sender=$row['SenderMemberID'];

  73.   $room=$row['RoomID'];



  77.  $sql2 =mysql_query("select MemberName from members where MemberID ='$sender'  ");
  78.  $sql1 =mysql_query("select RoomName,RoomLogo from rooms where RoomID ='$room' ");
  79.  while($row=mysql_fetch_array($sql2))
    {
        $mem =$row['MemberName'];
    }

  84.   while($rows=mysql_fetch_array($sql1))
    {   
        $Ro =$rows['RoomName'];
        $logo = $rows['RoomLogo'];
     }
  89.  ?>

Thanks alot