Re: passing return value from fucntion

Never mind I got it figured out. Had to set a global variable to make it visible. Thanks anyway -Will "Forest Liu" <oxware@gmail.com> kirjoitti viestissä:2f640a570503180005238eaa45@mail.gmail.com... I donot understand your problem clearly yet. I think there must be a caller and a callee (called function) when we talking about "passing arguments". That is to say, you write in your main program: ... echo calc($rs)."<br>"; ... and make sure there is a function named "calc": function calc($rightsid){ ... if(...){ } elseif ($rightsid == $oik6) { $res = ok; return $res; } elseif ($rightsid == $oik7) { $res = notok; return $res; } ... } then the result calculated in the function would be passed to the point you calling in your main program. Certainly,"ok" and "notok", I think they should be boolean, as "true" and "false". If you really write your code like that exactly, the two words "ok" and "notok" wont have any effective meaning. wish it helps to you. On Fri, 18 Mar 2005 09:33:13 +0200, William Stokes <kalles@operamail.com> wrote: > Hello, > Just simple question (I think?) > > How to pass return value from function to the main program? > > Here's example: > do some code in funtion to set the $res value: > > } elseif ($rightsid == $oik6) { > $res = ok; > return $res; > } elseif ($rightsid == $oik7) { > $res = notok; > return $res; > } > > How can I get to use the $res in the main program. The return $res; > doesn't > pass the variable to the main program right? > > Thanks > -Will > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php > > -- Sincerely, Forest Liu(???)