PHP: How to pass a function as an argument?

evdoxos · 03-02-2008, 02:48 AM

In PHP, what is the way to pass a function as an argument in an other function?

Dave · 03-02-2008, 06:05 AM

Well if you want to call a function you can use the call_user_func function. If you need to pass the function name as parameter you would do this...

Code:

<?php
function myFunction($callbackFunc)
{
    // Do whatever you got to here...

   // and then..
    call_user_func($callbackFunc);
}
?>

Hope this helps.

evdoxos · 03-02-2008, 02:00 PM

Thanks a lot. But my problem is more complicated:

<?php

class a
????: NamePros.com http://www.namepros.com/programming/439274-php-how-to-pass-function-parameter.html
{
public function f()
{
echo "af";
}
}

function func($g)
{
class b extends a
{
public function f()
{
echo "bf";
// I want to call function g from here
}
}
$b1 = new b();
$b1->f();
}

function g()
{
echo "g";
}

// From here I want to call function func with argument the funtion g

?>

Peter · 03-02-2008, 02:39 PM

why have you got a class within a function (class b is within the func function)?

This looks like very bad code design.

evdoxos · 03-02-2008, 03:21 PM

I need to call the following lines of code -the body of function func- from many locations in my application. So I use a function. Is it a bad idea?
(class a -not b- has a lot lines of code)
Also functions like g are more than 50!

class b extends a
{
public function f()
{
echo "bf";
// I want to call function g from here
}
}
$b1 = new b();
$b1->f();

03-02-2008, 02:48 AM	THREAD STARTER #1 (permalink)
evdoxos NamePros Member Join Date: Mar 2008 Posts: 46	PHP: How to pass a function as an argument? In PHP, what is the way to pass a function as an argument in an other function? Last edited by evdoxos; 03-02-2008 at 03:15 AM.

03-02-2008, 06:05 AM	#2 (permalink)
Dave Senior Member Join Date: Jun 2007 Location: NamePros.com Posts: 1,400	Well if you want to call a function you can use the call_user_func function. If you need to pass the function name as parameter you would do this... Code: <?php function myFunction($callbackFunc) { // Do whatever you got to here... // and then.. call_user_func($callbackFunc); } ?> Hope this helps.

03-02-2008, 02:39 PM	#4 (permalink)
Peter NamePros Expert Join Date: Nov 2003 Location: Scotland Posts: 5,069	why have you got a class within a function (class b is within the func function)? This looks like very bad code design. __________________ Manage your portfolio using my new Domain Portfolio Management script. Securing Your Domain Name From Theft

03-02-2008, 03:21 PM	THREAD STARTER #5 (permalink)
evdoxos NamePros Member Join Date: Mar 2008 Posts: 46	I need to call the following lines of code -the body of function func- from many locations in my application. So I use a function. Is it a bad idea? (class a -not b- has a lot lines of code) Also functions like g are more than 50! class b extends a { public function f() { echo "bf"; // I want to call function g from here } } $b1 = new b(); $b1->f(); Last edited by evdoxos; 03-02-2008 at 03:26 PM.

03-02-2008, 02:00 PM	THREAD STARTER #3 (permalink)
evdoxos NamePros Member Join Date: Mar 2008 Posts: 46	Thanks a lot. But my problem is more complicated: <?php class a ????: NamePros.com http://www.namepros.com/programming/439274-php-how-to-pass-function-parameter.html { public function f() { echo "af"; } } function func($g) { class b extends a { public function f() { echo "bf"; // I want to call function g from here } } $b1 = new b(); $b1->f(); } function g() { echo "g"; } // From here I want to call function func with argument the funtion g ?>

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