PHP: How to add a variable to a function name?

Question

I have a class with a callback. I pass a string to the class and the callback gets called with that string as the callback function. It works a little like this:

$obj->runafter( 'do_this' );

function do_this( $args ) {
    echo 'done';
}

What I am looking to do is run this inside of a loop and so that the function doesn't get written multiple times I want to add a variable to the function name. What I want to do is something like this:

for( $i=0;$i<=3;$i++ ) :
    $obj->runafter( 'do_this_' . $i );

    function do_this_{$i}( $args ) {
        echo 'done';
    }
endfor;

Any ideas on how I can accomplish this in PHP?

jszobody · Accepted Answer · 2016-03-15 15:46:22Z

6

I would pass the function in directly as a closure:

for($i=0; $i<=3; $i++) {
    $obj->runafter(function($args) use($i) {
        echo "$i is done";
    });
}

Note how you can use($i) to make this local variable available in your callback if needed.

Here is a working example: https://3v4l.org/QV66p

More info on callables and closures:

http://php.net/manual/en/language.types.callable.php

edited Mar 15, 2016 at 15:46

answered Mar 15, 2016 at 15:39

jszobody

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7 Comments

Jaquarh Over a year ago

Appreciated, that's actually madness x) Nice one!

jszobody Over a year ago

@KyleE4K just updated with a working example link as well

Jaquarh Over a year ago

I wonder if this is possible with C# (more strongly typed languages) - research time haha

Robert Over a year ago

@KyleE4K of course it is possible. In C# you can use Delegate

Aaron Over a year ago

Worked perfectly! Thanks!

|

Robert · Accepted Answer · 2016-03-16 15:17:46Z

2

You can assign function to variable and create closure also you can pass variable outside the scope of anonymous function via use keyword

$callback = function($args) use($i) {
    echo "$i is done";
};

for ($i = 0; $i <= 3; $i++) {
    $arg = "some argument";
    $obj->runafter($callback($arg));
}

Useful links:

edited Mar 16, 2016 at 15:17

answered Mar 15, 2016 at 15:46

Robert

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1 Comment

jszobody Over a year ago

Upvoted for showing how you can define the function as a closure first. Note you need a semi-colon at the end of the function definition, and you'll get Notice: Undefined variable: i. If notices are turned on.

Josh J · Accepted Answer · 2016-03-15 15:44:16Z

0

According to your question, you should define your function before the loop. Then you don't have to worry about it being redefined on each loop iteration.

function do_this( $args ) {
    echo 'done';
}

for( $i=0;$i<=3;$i++ ) :
    $obj->runafter( 'do_this' );
endfor;

answered Mar 15, 2016 at 15:44

Josh J

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