SQL從2個表中獲取信息並比較

Question

我有兩個表，我想比較並找到用戶名不符合的用戶，然後顯示未被跟蹤的用戶的用戶名。

Edit: 
Table 1: users 
username 

Table 2: follow 
username (the name of the user) 
followname (the name of the person the user follows) 

In Table 2 username and followname are never null. When a user follows another user then it is put into the list like so: 

Username   Followname 
derekshull   dvdowns 
derekshull   testuser 
dvdowns   testuser 
testuser   1511project 
derekshull   newuser 

在我腦子裏，我見表1獲得所有用戶名的列表（列名是「用戶名」，從表1），然後得到所有的用戶名的用戶（$用戶名）的列表不繼（表2中的列名是「followname」）。

如何讓它比較兩個列表並且不顯示用戶已經遵循的用戶名？

這就是我現在所擁有的。出於某種原因，它正在顯示我遵循並且不遵循的用戶。

$alluserslookup = mysql_query("SELECT * FROM users WHERE username!='$username'"); 
$thefollowerslookup = mysql_query("SELECT * FROM follow WHERE username='$username'"); 

while ($followersrow = mysql_fetch_assoc($thefollowerslookup)) { 
$afollower = $followersrow['followname']; 

while ($allusersrow = mysql_fetch_assoc($alluserslookup)) { 
$allusers = $allusersrow['username']; 

if ($afollower != $allusers) { 
echo "<a href='viewprofile.php?viewusername=$allusers'>$allusers</a><br>"; 
} 
} 
} 

Answer 1

（更新）嘗試：

select u.* from users u 
left join follow f on u.username = f.followname and f.username = 'derekshull' 
where f.followname is null 

SQLFiddle here。