How to call PHP function from string stored in a V

Question

I need to be able to call a function, but the function name is stored in a variable, is this possible? e.g:

function foo ()
{
  //code here
}

function bar ()
{
  //code here
}

$functionName = "foo";
// i need to call the function based on what is $functionName

Anyhelp would be great.

Thanks!

Answer 1

For the sake of completeness, you can also use eval():

$functionName = "foo()";
eval($functionName);

However, call_user_func() is the proper way.

Answer 2

In case someone else is brought here by google because they were trying to use a variable for a method within a class, the below is a code sample which will actually work. None of the above worked for my situation. The key difference is the & in the declaration of $c = & new... and &$c being passed in call_user_func.

My specific case is when implementing someone's code having to do with colors and two member methods lighten() and darken() from the csscolor.php class. For whatever reason, I wanted to have the same code be able to call lighten or darken rather than select it out with logic. This may be the result of my stubbornness to not just use if-else or to change the code calling this method.

$lightdark="lighten"; // or optionally can be darken
$color="fcc";   // a hex color
$percent=0.15;  
include_once("csscolor.php");
$c = & new CSS_Color($color);
$rtn=call_user_func( array(&$c,$lightdark),$color,$percent);

Note that trying anything with $c->{...} didn't work. Upon perusing the reader-contributed content at the bottom of php.net's page on call_user_func, I was able to piece together the above. Also, note that $params as an array didn't work for me:

// This doesn't work:
$params=Array($color,$percent);
$rtn=call_user_func( array(&$c,$lightdark),$params);

This above attempt would give a warning about the method expecting a 2nd argument (percent).

Answer 3

I dont know why u have to use that, doesnt sound so good to me at all, but if there are only a small amount of functions, you could use a if/elseif construct. I dont know if a direct solution is possible.

something like $foo = "bar"; $test = "foo"; echo $$test;

should return bar, you can try around but i dont think this will work for functions

Answer 4

A few years late, but this is the best manner now imho:

$x = (new ReflectionFunction("foo"))->getClosure();
$x();

Answer 5

Dynamic function names and namespaces

Just to add a point about dynamic function names when using namespaces.

If you're using namespaces, the following won't work except if your function is in the global namespace:

namespace greetings;

function hello()
{
    // do something
}

$myvar = "hello";
$myvar(); // interpreted as "\hello();"

What to do?

You have to use call_user_func() instead:

// if hello() is in the current namespace
call_user_func(__NAMESPACE__.'\\'.$myvar);

// if hello() is in another namespace
call_user_func('mynamespace\\'.$myvar);

Answer 6

$functionName() or call_user_func($functionName)