What should I depend on if php functions are not called immediately?

Question

I've tried regular functions and call_user_function, both which do not finish executing before they get to the next line after a function call. What do people use instead? Should I just include a separate php file for every "function" I want to run?

edit:

$IDNum = 0;


$restartButtonIDName;

$playButtonIDName;

$audioPlayerIDName;

$MP3AudioSourceIDName;
$OGGAudioSourceIDName;

$resultingTextIDName;

$currentTimeIDName;

$checkOffsetIDName;



call_user_func('setIDNames');



function setIDNames() {

    echo "called setIDNames <br />";

    call_user_func('incIDNums');

    $restartButtonIDName = "restartButton".$IDNum;

    $playButtonIDName = "playButton".$IDNum;

    $audioPlayerIDName = "audioPlayer".$IDNum;

    $MP3AudioSourceIDName = "MP3AudioSource".$IDNum;
    $OGGAudioSourceIDName = "OGGAudioSource".$IDNum;

    $resultingTextIDName = "resultingText".$IDNum;

    $currentTimeIDName = "currentTime".$IDNum;

    $checkOffsetIDName = "checkOffset".$IDNum;


}

function incIDNums(){

    echo "called setIDNums <br />";

    $IDNum += 1;

}

echo $restartButtonIDName." test<br />"; // echos "test" not the resulting name


?>

You need to provide more information. Perhaps a code sample that demonstrates your issue? — Andy, Jul 02 '14 at 19:11
What are you talking about? Most PHP functions are synchronous, the next line doesn't execute until they're finished. — Barmar, Jul 02 '14 at 19:11
I have the variables defined first outside of the functions. — Chewie The Chorkie, Jul 02 '14 at 19:17
Sourijrou I call them with call_user_func. I also tried the regular way of calling a function like setIDNames(); — Chewie The Chorkie, Jul 02 '14 at 19:18
Why do you have all those lines that just have a variable name on them? That doesn't do anything. — Barmar, Jul 02 '14 at 19:19
Why do the functions need to return anything? I have the variables placed outside so they can be changed by the functions. — Chewie The Chorkie, Jul 02 '14 at 19:20
Unlike Javascript, the fact that you define the variables outside of the function does not make them global when using them inside the function. You have to specify that you want the global instances. — Patrick Q, Jul 02 '14 at 19:21

Barmar · Accepted Answer · 2014-07-02T19:28:53.277

Variables accessed inside a function are not the same as variables outside the function, or in a different function, unless you declare the variables with a global statement inside the function.

$IDNum = 0;
incIDNums();
echo $IDNum; // Will echo 1

function incIDNums(){
    global $IDNum;

    echo "called setIDNums <br />";
    $IDNum += 1;

}

It's generally considered poor style to depend heavily on global variables, as it impairs the generality of the functions. The function should get its input via parameters, and sends the results as a value using return. If you need to return multiple results, you can package them into an array, or use reference parameters that are updated in place.

score 1 · Answer 2 · answered Jul 02 '14 at 19:26

Probably the problem you have is due to the scope of variables ,please note the lifetime of variables inside a function is only until function is active to make changes effect to global variables which are outside of function you need to make them global inside function like:

function setIDNames() {
    global $restartButtonIDName ,$playButtonIDName ,$audioPlayerIDName ...;//All variables you wanna use saperated by comma.
    //Your code
}

What should I depend on if php functions are not called immediately?

2 Answers2