PHP - How to split array into variables and pass to a function?

Question

There are a few functions with different number of parameters in PHP. I have no idea which function is going to be called, but the function and its parameter is passed to the calling function as array like this

function func1($a,$b){...}
function func2($a){...}
$calls  = [func1=>[arg1, arg2], func2=>[arg1]]

I need to call each function with its parameters. I don't know how to pass the parameters as distinct variables. This my code

$func_names = array_keys($calls);
$i = 0;
foreach($calls as $call){
    $func_names[$i]($calls[$func_names[$i++]]);
    //$func_names[$i]($call); same as above line
}

In each iteration array of arguments of each function is passed to the function not each item of the array separately. How can I solve this problem?

thanks

Is this helpfull: php.net/manual/en/function.func-get-args.php ? — Martijn
– Martijn, Commented Jan 7, 2019 at 9:04
@Martijn I didn't try that. Let me try and will let you on it — M a m a D
– M a m a D, Commented Jan 7, 2019 at 9:06

Yehonatan · Accepted Answer · 2019-01-07 09:07:19Z

2

Use call_user_func_array

mixed call_user_func_array ( callable $callback , array $param_arr )

Example from the linked PHP manual -

<?php
function foobar($arg, $arg2) {
    echo __FUNCTION__, " got $arg and $arg2\n";
}
class foo {
    function bar($arg, $arg2) {
        echo __METHOD__, " got $arg and $arg2\n";
    }
}


// Call the foobar() function with 2 arguments
call_user_func_array("foobar", array("one", "two"));

// Call the $foo->bar() method with 2 arguments
$foo = new foo;
call_user_func_array(array($foo, "bar"), array("three", "four"));
?>

answered Jan 7, 2019 at 9:07

Yehonatan

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1 Comment

M a m a D Over a year ago

Very perfect and clean answer. Thank you

Progrock · Accepted Answer · 2019-01-07 09:47:58Z

2

<?php

function a($foo) {
    echo $foo, "\n";
}
function b($bar, $baz) {
    echo $bar, ' and ', $baz, "\n";
}

$calls  = ['a'=>['apples'], 'b'=>['bananas','banjos']];

foreach($calls as $k => $v) $k(...$v);

Output:

apples
bananas and banjos

answered Jan 7, 2019 at 9:47

Progrock

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