Re: Random pick
| From: | Karl DeSaulniers | Date: | Mon, 25 Jan 2010 20:33:44 +0000 |
| Subject: | Re: Random pick | ||
| References: | 1 2 3 4 5 6 7 8 9 10 | Groups: | php.db |
| Request: | Send a blank email to php-db+get-46802@lists.php.net to get a copy of this message | ||
Yes, $req_user_level is an int between 0 and 9.
Yes, the other code worked great.
Thank you for your response though.
That is definitely a good way to cross check.
Thanks again for your responses.
Best,
Karl
On Jan 25, 2010, at 2:24 PM, Peter Beckman wrote:
On Mon, 25 Jan 2010, Karl DeSaulniers wrote:Karl DeSaulniers Design Drumm http://designdrumm.comThank 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@angryox.com http://www.angryox.com/ -----------------------------------------------------------------------------PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php