Re: Function -> Sending URL's

Hey Dan You see in this example - the value of $company is coming from the script and not the function - I am trying to find a way of telling the function which variable I want to use, but the value for that variable to come from the function for exmaple: I want to tell the function - 'here is the name of the link, here are the variables to attach onto the end of it - but work out the values yourself!' Does that makes sense - I can't find a way to do that. Thanks, Abe ----- Original Message ----- From: "DanO" <dano@informano.net> To: "Abe" <abe@fish.tm>; <php-general@lists.php.net> Sent: Thursday, January 11, 2001 9:41 PM Subject: RE: [PHP] Function -> Sending URL's > > sorry, but i didn't fully understand what you are trying to do. > > you can also pass 2 variables to the function: > > <? > function testPassVar($url,$param) { > echo "<a href="; > echo $url; > echo "?name="; > echo $param; > echo ">Click Here</a>"; > } > > $script = 'asdfasdf.php3'; > $company = 'TEST'; > testPassVar($script,$company); > ?> > > > > -----Original Message----- > From: DanO [mailto:dano@informano.net] > Sent: Thursday, January 11, 2001 1:35 PM > To: Abe; php-general@lists.php.net > Subject: RE: [PHP] Function -> Sending URL's > > > > you are assigning the variable within the function, then calling the > function with the variable. > > this is a variable scoping issue. > > try this: > > <? > function testPassVar($url) { > echo "<a href=\"$url\">Click</a>"; > } > > $company = "TEST"; > $url = 'asdfasdf.php3?name='.$company; > testPassVar($url); > ?> > > > > -----Original Message----- > From: Abe [mailto:abe@fish.tm] > Sent: Thursday, January 11, 2001 1:35 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 > > > -- > 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 >