RE: [PHP] Coding hints?

Its a cast to an int where does $sb get its value from? You can put anything in $sb, by default its a string. Eg If you have a2 in $sb then $arr = file('sb' . $sb . '.src'); will be sbs2.src -----Original Message----- From: Gavster [mailto:gavster@offbeatmedia.com] Sent: November 8, 2000 10:34 AM To: php-general@lists.php.net Subject: [PHP] Coding hints? Hi Guys, Sorry to post with something trivial, but I can't seem to find anything on this: I'm modifying some code to recognize a certain filename scheme, but I've run across something I've never seen before. Here's the code: $arr = file('sb' .(int)$sb. '.src'); The filename it's looking for is in the form sbN.src (where N = an integer). The $sb variable is an integer passed via the url. What I don't get is the (int)$sb part - what is the (int) doing there? Is this some sort of funky regex? Or is it casting the $sb string into an int? I must admit, I'm at a loss . . . What I'm ultimately trying to do is to get the code to recognize files where $sb is not only a number, but is also a number with a single char appended to it such as [a-z]. Can't this be done with a straightforward regex? Any pointers would be greatly appreciated . . . =0) -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: php-general-unsubscribe@lists.php.net For additional commands, e-mail: php-general-help@lists.php.net To contact the list administrators, e-mail: php-list-admin@lists.php.net