RE: [PHP] Function -> Sending URL's

There's a couple points here: 1: echo isn't what you want to do. You want to RETURN the value from you function. Try this: <? function testPassVar($url) { $name = "TEST"; echo "<a href=\"$url\">Click</a>"; return "<a href=\"$url?name=$name\">Click</a>"; } $url = 'asdfasdf.php3'; $newURL = testPassVar($url); echo($newURL); echo "This is a $name"; ?> 2: in the last line, you try to echo $name. $name is defined within the function testPassVar() and goes out of scope when that function is done At this point in your script, $name does not exist. Hope this helps, Cal http://www.calevans.com -----Original Message----- From: Abe [mailto:abe@fish.tm] Sent: Thursday, January 11, 2001 3:35 PM To: php-general@lists.php.net Subject: [PHP] Function -> Sending URL's Hey there, this is a strange one - I want to send a URL to a function that includes varibles. Those variables should be taken from within the function - as in the example below the link I want is: asdfasdf.php3?name=TEST , but that is not what I get as you can see. Does anybody know a way around this - The example is simpler than what I am actually doing and the value of $company must come from the variable. Thanks, Abe <? function testPassVar($url) { $name = "TEST"; echo "<a href=\"$url\">Click</a>"; } $url = 'asdfasdf.php3?name=$company'; testPassVar($url); echo "This is a $name"; ?> -- 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