How to pass variable number of arguments to a PHP function

Question

I have a PHP function that takes a variable number of arguments (using func_num_args() and func_get_args()), but the number of arguments I want to pass the function depends on the length of an array. Is there a way to call a PHP function with a variable number of arguments?

Answer 1

If you have your arguments in an array, you might be interested by the call_user_func_array function.

If the number of arguments you want to pass depends on the length of an array, it probably means you can pack them into an array themselves -- and use that one for the second parameter of call_user_func_array.

Elements of that array you pass will then be received by your function as distinct parameters.

For instance, if you have this function :

function test() {
  var_dump(func_num_args());
  var_dump(func_get_args());
}

You can pack your parameters into an array, like this :

$params = array(
  10,
  'glop',
  'test',
);

And, then, call the function :

call_user_func_array('test', $params);

This code will the output :

int 3

array
  0 => int 10
  1 => string 'glop' (length=4)
  2 => string 'test' (length=4)

ie, 3 parameters ; exactly like iof the function was called this way :

test(10, 'glop', 'test');

Answer 2

This is now possible with PHP 5.6.x, using the ... operator (also known as splat operator in some languages):

Example:

function addDateIntervalsToDateTime( DateTime $dt, DateInterval ...$intervals )
{
    foreach ( $intervals as $interval ) {
        $dt->add( $interval );
    }
    return $dt;
}

addDateIntervaslToDateTime( new DateTime, new DateInterval( 'P1D' ), 
        new DateInterval( 'P4D' ), new DateInterval( 'P10D' ) );

Answer 3

In a new Php 5.6, you can use ... operator instead of using func_get_args().

So, using this, you can get all the parameters you pass:

function manyVars(...$params) {
   var_dump($params);
}

Answer 4

Since PHP 5.6, a variable argument list can be specified with the ... operator.

function do_something($first, ...$all_the_others)
{
    var_dump($first);
    var_dump($all_the_others);
}

do_something('this goes in first', 2, 3, 4, 5);

#> string(18) "this goes in first"
#>
#> array(4) {
#>   [0]=>
#>   int(2)
#>   [1]=>
#>   int(3)
#>   [2]=>
#>   int(4)
#>   [3]=>
#>   int(5)
#> }

As you can see, the ... operator collects the variable list of arguments in an array.

If you need to pass the variable arguments to another function, the ... can still help you.

function do_something($first, ...$all_the_others)
{
    do_something_else($first, ...$all_the_others);
    // Which is translated to:
    // do_something_else('this goes in first', 2, 3, 4, 5);
}

---

Since PHP 7, the variable list of arguments can be forced to be all of the same type too.

function do_something($first, int ...$all_the_others) { /**/ }

Answer 5

For those looking for a way to do this with $object->method:

call_user_func_array(array($object, 'method_name'), $array);

I was successful with this in a construct function that calls a variable method_name with variable parameters.

Answer 6

You can just call it.

function test(){        
     print_r(func_get_args());
}

test("blah");
test("blah","blah");

Output:

Array ( [0] => blah ) Array ( [0] => blah [1] => blah )

Answer 7

I'm surprised nobody here has mentioned simply passing and extracting an array. E.g:

function add($arr){
    extract($arr, EXTR_REFS);
    return $one+$two;
}
$one = 1;
$two = 2;
echo add(compact('one', 'two')); // 3

Of course, this does not provide argument validation. For that, anyone can use my expect function: https://gist.github.com/iautomation/8063fc78e9508ed427d5

Answer 8

An old question, I know, however, none of the answers here really do a good job of simply answer the question.

I just played around with php and the solution looks like this:

function myFunction($requiredArgument, $optionalArgument = "default"){
   echo $requiredArgument . $optionalArgument;
}

This function can do two things:

If its called with only the required parameter: myFunction("Hi") It will print "Hi default"

But if it is called with the optional parameter: myFunction("Hi","me") It will print "Hi me";

I hope this helps anyone who is looking for this down the road.

Answer 9

Here is a solution using the magic method __invoke

(Available since php 5.3)

class Foo {
    public function __invoke($method=null, $args=[]){
        if($method){
            return call_user_func_array([$this, $method], $args);
        }
        return false;
    }

    public function methodName($arg1, $arg2, $arg3){

    }
}

From inside same class:

$this('methodName', ['arg1', 'arg2', 'arg3']);

From an instance of an object:

$obj = new Foo;
$obj('methodName', ['arg1', 'arg2', 'arg3'])