[SOLVED] non objects/undefined variable.

[AsiaUnderworld]

<?


$id = secure($_SESSION['user_id']);
$id = $id[0];

        $sql = "SELECT * FROM user_stats WHERE user_id='$id'";
	$result = mysql_query($sql); 
	$result2 = mysql_fetch_assoc($result);

	$firstcheck = "SELECT ticketprice FROM airport WHERE id='1' ";
	$firstcheck1 = "SELECT ticketprice FROM airport WHERE id='2' ";
	$firstcheck2 = "SELECT ticketprice FROM airport WHERE id='3' ";
	$firstcheck3 = "SELECT ticketprice FROM airport WHERE id='4' ";

	$price = mysql_query($firstcheck) or die(mysql_error());
	$price1 = mysql_query($firstcheck1)  or die(mysql_error());
	$price2 = mysql_query($firstcheck2)  or die(mysql_error());
	$price3 = mysql_query($firstcheck3)  or die(mysql_error());



if(isset($_POST['travel']))
if (isset($_POST['radio']))
{
$radio=strip_tags($_POST['radio']);

if (!ereg('[^0-9]',$radio))
{
	if ($radio == "1")
	{
		$costs = $price ;
		$to = "Chicago";
	}
	elseif ($radio == "2")
	{
		$costs = $price1;
		$to = "Detroit";
	}
	elseif ($radio == "3")
	{
		$costs = $price2 ;
		$to = "New York";
	}
	elseif ($radio == "4")
	{
		$costs = $price3 ;
		$to = "Washington";
	}
}
if ($fetch->lasttravel > time())
	echo "You cannot travel for another ".maketime($result2->lasttravel)." "; 
elseif ($costs > $result2->money)
	echo " You don't have enough money for this ticket" ; 
}
elseif ($costs <= $result2->money)
{
}
elseif ($to == $result2->location_id)
	echo " You are already in this city. " ;
{
}
if ($to != $result2->location_id)
{
$flightcosts = $result2->money - $costs;
$now = time() + 3600;

mysql_query("UPDATE user_stats SET location_id='$to', money='$flightcosts' , lasttravel='$now'  WHERE user_id='$id'");
}
?>

 

The page says

 

15:17 pm:

Notice: Trying to get property of non-object in /home/djwdesi/public_html/gangster/airport.php on line 80

15:17 pm:

Notice: Undefined variable: to in /home/djwdesi/public_html/gangster/airport.php on line 80

 

Those lines are

 

}

if ($to != $result2->location_id)

{

 

When i click the submitbutton to travel it still does nothing, whats wrong?

 

thanks

[flyhoney]

Add this and see what you get:

 

<?php
$result = mysql_query($sql) or die('Invalid query: ' . mysql_error() . "\n"); 
?>

[Maq]

Don't you mean

 

   elseif ($to == $result2->location_id)
{
      echo " You are already in this city. " ;
}

 

Not:

 

   elseif ($to == $result2->location_id)
      echo " You are already in this city. " ;
{
}

[AsiaUnderworld]

tried both and nothing changed

[flyhoney]

OH, duh.

 

if ($to != $result2->location_id)

 

should be

 

if ($to != $result2['location_id'])

[AsiaUnderworld]

that doesnt do anything, just makes more errors

[flyhoney]

Well, when you call mysql_fetch_assoc() on a mysql result it returns an associative array, not an object.  Hence the "assoc".  You are referencing an associative array as an object which is why you are getting the errors.  Either use mysql_fetch_object() or start referencing $result2 as an associative array.