SQL从2个表中获取信息并比较

Question

我有两个表，我想比较并找到用户名不符合的用户，然后显示未被跟踪的用户的用户名。

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 

在我脑子里，我见表1获得所有用户名的列表（列名是“用户名”，从表1），然后得到所有的用户名的用户（$用户名）的列表不继（表2中的列名是“followname”）。

如何让它比较两个列表并且不显示用户已经遵循的用户名？

这就是我现在所拥有的。出于某种原因，它正在显示我遵循并且不遵循的用户。

$alluserslookup = mysql_query("SELECT * FROM users WHERE username!='$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）尝试：

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。