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我有 2 个表要比较并找到用户名未关注的用户，然后显示未关注的人的用户名。

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

在我的脑海中，我看到表 1 获取了所有用户名的列表（列名是表 1 中的“用户名”），然后获取了用户 ($username) 未关注的所有用户名的列表（列名是“followname”来自表 2)。

如何让它比较两个列表而不显示用户已经关注的用户名？

这就是我现在所拥有的。出于某种原因，它正在显示我关注和不关注的用户。

$alluserslookup = mysql_query("SELECT * FROM users WHERE username!='$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>";
}
}
}
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1 回答 1

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在这里。

于 2013-04-24T16:01:04.750 回答