Re: [PHP-DB] Random pick

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
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Peter Beckman                                                  Internet Guy
[email protected]                                 http://www.angryox.com/
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