Re: calling functions to form menus

Sounds like your functions look like this: Menu() { echo $link1; echo $link2; } when they need to look like this: Menu() { $ret .= $link1; $ret .= $link2; return $ret; } Basically assign EVERYTHING that is supposed to be "echo"'d and then return that into your next function. --Joe On Mon, Dec 04, 2000 at 02:21:17PM -0700, webmaster@ipo.vpaa.asu.edu wrote: > Greetings, > > Got a tiny problem that I'm unable to figure out. In one file I've > written two functions, one which forms the table layout for the page, > and another which forms the links that make up the menu side of the > table. > > In the page that calls the functions I've tried assigning the output of > the Menu() function to a variable $menu and insert it into the main > page. > > require("FormatPage.php"); > > $menu = Menu(); > $title ="stuff"; > $header="more stuff"; > > MainPage($title, $header, $menu); > > > The menu does not get placed inside the table of MainPage, instead it's > formed above the table. Other than that the functions work fine. > > Is there any way to assign the output of a function to a single variable > and place it inside the table where it's supposed to be? > > Thanks. > > -- > 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 ----- Joe Stump, PHP Hacker joestump98@yahoo.com http://www.miester.org/ Need a programmer? http://www.google.com/search?q=%22Joe+Stump%22 ***************************************************************** * RENT THIS SPACE - SEEN BY THOUSANDS EACH DAY! * * Current Sponsor: http://www.smackdown.com * *****************************************************************