php call_user_func_array doesn't work

Question

first time trying to deal with this call_user_func_array, but something isn't working, since I get no response from the function, what can I be missing?

function _a_($id, $text) {          
    if($id == 'a') {
    _b_();
    if(substr($text, 0, 8) == "{source}") {
        $campos = substr_replace($text, '', 0, 8);
        $campos = substr($campos, 0, -9);
        $funcao = explode(";", $campos);
        print_r($funcao);
        call_user_func_array($funcao[0], $funcao[1]);
    }
    } else {
        echo $text."<br>";
    }
}
function _b_() {
    echo "b was fired<br>";
}
function _c_($some_text) {
    echo "received a call<br>";
    echo "inside function c: ".$some_text."<br>";
}
_a_("a", "{source}_c_;ola{/source}");

Enable your diagnostics, you get a big fat error message telling you exactly what's wrong. — Kerrek SB
– Kerrek SB, Commented Oct 10, 2011 at 15:26
Array ( [0] => c [1] => ola ) It now works with the call_user_func_array($funcao[0], array($funcao[1])); — Pluda
– Pluda, Commented Oct 10, 2011 at 15:39

cuttinger · Accepted Answer · 2011-10-10 15:29:40Z

4

call_user_func_array($funcao[0], $funcao[1]);

=>

call_user_func_array($funcao[0], array($funcao[1]));

answered Oct 10, 2011 at 15:29

cuttinger

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Comments

Kerrek SB · Accepted Answer · 2011-10-10 15:27:30Z

4

Say either this:

call_user_func($funcao[0], $funcao[1]);

Or this:

call_user_func_array($funcao[0], array($funcao[1]));

The latter form is only useful if you need to pass the arguments by reference; see the documentation for details.

answered Oct 10, 2011 at 15:27

Kerrek SB

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Comments

Alessandro Desantis · Accepted Answer · 2011-10-10 15:28:16Z

3

call_user_func_array() expects the second parameter to be an array. Use call_user_func if you know the number of parameters.

answered Oct 10, 2011 at 15:28

Alessandro Desantis

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2 Comments

Pluda Over a year ago

can't do, I want this to be dynamic

Alessandro Desantis Over a year ago

Then use call_user_func_array().

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