为什么我的SQL插入工作？

I know questions like this exist all over, but I can't see what's wrong with my code. The DB works and I run an update query just fine earlier on in the code.

Query 1:

mysql_query("INSERT INTO `login_history` (`memberid`, `ip`, `host`, `location`, `status`, `date`) VALUES ('".$login."', '".$ip."', '".$details->hostname."', '".$loc."', 'success', NOW()");

Query 2:

mysql_query("INSERT INTO `login_history` (`memberid`, `ip`, `host`, `location`, `status`, `date`) VALUES ('".$login."', '".$ip."', '".$details->hostname."', '".$loc."', 'failure', NOW()");

Here is an echo of the string as requested:

INSERT INTO `login_history` (`memberid`, `ip`, `host`, `location`, `status`, `date`) VALUES ('AAL', '**.60.**.**', 'c-174-**-**-**.hsd1.**.comcast.net', 'Town, US', 'success', NOW()

Looks like you are missing the final closing ). Also use prepared statements. Much cleaner and safer. Here is a quick example of what a Prepared Statement would look like (adapted from here) (you wold also need to make other changes to your PHP script to start using them)

$stmt = $mysqli->prepare("INSERT INTO `login_history` (`memberid`, `ip`, `host`, `location`, `status`, `date`) VALUES (?, ?, ?, ?, ?, NOW())")
$stmt->bind_param('sssss', $login, $ip, $details->hostname, $loc, 'failure');
$stmt->execute()