PHP - Pass array as variable-length argument list

Question

I have really simple problem in my PHP script. There is a function defined which takes variable length argument list:

function foo() {
  // func_get_args() and similar stuff here
}

When I call it like this, it works just fine:

foo("hello", "world");

However, I have my variables in the array and I need to pass them "separately" as single arguments to the function. For example:

$my_args = array("hello", "world");
foo(do_some_stuff($my_args));

Is there any do_some_stuff function which splits the arguments for me so I can pass them to the function?

possible duplicate of Passing an Array as Arguments, not an Array, in PHP — DanMan
– DanMan, Commented Feb 18, 2014 at 14:31
Use this answer. This is much more accurate. stackoverflow.com/a/34049110/2147023 — theBuzzyCoder
– theBuzzyCoder, Commented Jun 23, 2018 at 9:14
Does this answer your question? unpacking an array of arguments in php — Anderson Green
– Anderson Green, Commented Aug 16, 2021 at 14:08

Aubin · Accepted Answer · 2014-12-26 15:36:42Z

17

Use

ReflectionFunction::invokeArgs(array $args)

or

call_user_func_array( callback $callback, array $param_arr)

edited Dec 26, 2014 at 15:36

Aubin

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answered Jan 28, 2012 at 12:57

Crozin

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1 Comment

dfsq Over a year ago

+1 for Reflections. Although it's like an overkill here, but it's always cool to deal with them!

dfsq · Accepted Answer · 2012-01-28 12:57:35Z

6

Well you need call_user_func_array

call_user_func_array('foo', $my_args);

http://php.net/manual/en/function.call-user-func-array.php

answered Jan 28, 2012 at 12:57

dfsq

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1 Comment

Pavel S. Over a year ago

Thanks for the answer, it's exactly what I need. However, I can accept only one answer.

Peter O'Callaghan · Accepted Answer · 2012-01-28 12:57:02Z

3

Sounds to me like you are looking for call_user_func_array.

answered Jan 28, 2012 at 12:57

Peter O'Callaghan

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2 Comments

Pavel S. Over a year ago

Thanks for the answer, it's exactly what I need. However, I can accept only one answer.

Peter O'Callaghan Over a year ago

Just accept the one that either gives the most required detail, or the earliest one :)

Community · Accepted Answer · 2023-11-17 19:59:20Z

3

You are searching for call_user_func_array().

https://www.php.net/manual/en/function.call-user-func-array.php

Usage:

$my_args = array("hello", "world");
call_user_func_array('foo', $my_args);

// Equivalent to:
foo("hello", "world");

edited Nov 17, 2023 at 19:59

CommunityBot

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answered Jan 28, 2012 at 12:59

lorenzo-s

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1 Comment

StormByte Over a year ago

This is far better than to use Reflection in my opinion

Community · Accepted Answer · 2017-05-23 12:24:54Z

0

http://www.php.net/manual/en/functions.arguments.php#functions.variable-arg-list

Isn't this what you want?

edit ah... ok... how about this: Passing an Array as Arguments, not an Array, in PHP

edited May 23, 2017 at 12:24

CommunityBot

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answered Jan 28, 2012 at 12:54

Yevgeny Simkin

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1 Comment

Pavel S. Over a year ago

Nope. I have already read this. All the functions are supposed to work inside my "foo" function. I need to split the arguments before they are passed to the function.

Arjan · Accepted Answer · 2012-01-28 12:58:16Z

0

If you can change the code of foo() it should be easy to solve this in just one place.

function foo()
{
    $args = func_get_args();
    if(count($args) == 1 && is_array($args[0]))
    {
        $args = $args[0]
    }
    // use $args as normal
}

answered Jan 28, 2012 at 12:58

Arjan

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2 Comments

lorenzo-s Over a year ago

This not works if he wants to pass an array (for real) as first argument.

Arjan Over a year ago

No, it does not work if there's only one argument and that argument is supposed to be an array. If there's more than one argument then they're all left alone.

Morten Haggren · Accepted Answer · 2021-06-28 11:48:28Z

0

I know it's an old question but it still comes up as the first search result - so here is an easier way;

<?php
function add(... $numbers) {
    $result=0;
    foreach($numbers as $number){
      $result+=intval($number);
    }
    return $result;
}

echo add(...[1, 2])."\n";

$a = [1, 2];
echo add(...$a);
?>

Source: https://www.php.net/manual/en/functions.arguments.php#example-142

answered Jun 28, 2021 at 11:48

Morten Haggren

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Comments

Pheonix · Accepted Answer · 2012-01-28 13:00:19Z

-1

This solution is not recommended at all, but just showing a possibility :

Using eval

eval ( "foo('" . implode("', '", $args_array) . "' )" );

answered Jan 28, 2012 at 13:00

Pheonix

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1 Comment

Crozin Over a year ago

I know it's only "showing a possibility", but still... it's eval. ;)

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