On Mon, 25 Jan 2010, Karl DeSaulniers wrote:
Thank you for this as well.
Question? What part is "in_array" playing?
Is it comparing $req_user_level to array()?
Because the text "Guest", etc.. is not in $req_user_level on the database.
In other words, is it checking the value of $req_user_level to see if "Guest"
is in it?
Sorry, I missed the question mark. in_array isn't appropriate here. The
previous poster has it right, assuming $req_user_level is an integer of
0..9.
$levels = array("Guest", "Regular User", 'Intl. User', ...)
// the array is 0 , 1 , 2 , ...)
Going a little further:
if (!empty($levels[$req_user_level])) { // is both set and doesn't evaluate to false
echo "The user is a {$levels[$req_user_level]}.\n";
} else {
// The $req_user_level was not a valid level.
echo "The returned req_user_level was not valid.\n";
}
Which would output, if $req_user_level was 1 (one):
The user is a Regular User.
Then you know you have a valid user level. Careful though -- sometimes 0
will be returned on a failure, depending on your SQL.
Beckman
---------------------------------------------------------------------------
Peter Beckman Internet Guy
beckman@xxxxxxxxxxx http://www.angryox.com/
---------------------------------------------------------------------------
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
