Need help PLEASE

[advo]

ok i have this warning showing up

 

Warning: in_array() [function.in-array]: Wrong datatype for second argument in /home/lalala/public_html/attack.php on line 72

 

the code i use

//same ip resrictions
$result = mysql_query("SELECT ip FROM ip_history WHERE user_id='$target[id]'");


while($row = mysql_fetch_array($result))
{
$sendtouser[]=$row['ip'];
}


$result = mysql_query("SELECT ip FROM ip_history WHERE user_id='$logged[id]'");

reconnect();

while($row = mysql_fetch_array($result))
{
if(in_array($row['ip'], $sendtouser))
{
	show_error("You cannot attack a user with the same IP.");
	break;
}
}

 

basically what i can gather is there is no ip for the user being attacked

i want to know how to stop this error tomming up if the $target[id] user has no ip in ip history any solution or fix to this would be greatly appreciated.

 

[Adam]

The error is trying to tell you the second parameter isn't an array (which it needs to be). At a guess I'd say that the first query isn't returning any results and so the $sendtouser array isnever being populated. Declare the array before you loop through the results, so even if it's empty, it's still an array when it comes to your in_array() condition:

 

$sendtouser = array();

while($row = mysql_fetch_array($result))
{
   $sendtouser[]=$row['ip'];
}

[advo]

ok thankyou very much i will try this now

 

tried it, you are a god i knew it be simple just so late my brains not with it atm thanks alot