这个MySQL查询有什么问题？

Question

It's 12:30am and I have been coding for 9 hours straight. I really need to get this project done, but MySQL is messing with my deadline. Could you examine this snippet for me and see if you can find out what is wrong?

PHP/MySQL Query

$q = $this->db->query("SELECT * FROM bans WHERE ip='".$ip."'");

Keeps returning the following error...

MYSQL Error [Oct 6th, 2010 11:31pm CDT]
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '* FROM bans WHERE ip='206.53.90.231'' at line 1 (1064)

I do not see anything wrong with the query. I've even tried different methods of including the variable $ip but with no avail.

EDIT:
Just to add in here, the ip column in my database is a varchar(255).

EDIT 2:
Here is the whole affected code. Keep in mind that this is all in a class. If I'm missing something, let me know.

Line from another Function

if($this->isBanned($_SERVER['REMOTE_ADDR'])===true) { return json_encode(array('error'=>'You are banned from this ShoutBox.')); }

Affected Function

function isBanned($ip) {
    $q = $this->db->query("SELECT * FROM bans WHERE ip='".$ip."'"); $num = $this->db->affected_rows;
    if($num>0) { $row = $this->db->fetch_array($q); if(($row['expires'] < time()) && ($row['expires'] !== 0)) { $this->unbanUser($ip,'internal'); return false; } return true; } return false;
}

unbanUser function

function unbanUser($ip,$t='box') {
    $q = $this->db->query("SELECT * FROM bans WHERE ip='".$ip."'"); $num = $this->db->affected_rows; if($num>0) { $q = $this->db->query("DELETE * FROM bans WHERE ip='".$ip."'"); 
    return (($t=='box') ? json_encode(array('status'=>'removed')) : true); } else { return (($t=='box') ? json_encode(array('error'=>'Unable to locate the user.')) : true); }
}

Answer 1

I think it may be It is your DELETE statement which is causing the error.

Remove the * after the DELETE and it should be fine.

Answer 2

试试这个：

$q = $this->db->query('SELECT * FROM bans WHERE ip="' . $ip . '"');

Answer 3

检查您是否使用'字符或'字符（最后一个是重音）