Callback in function (PHP) is not working

Question

When I execute following code I am getting this error. Why is that? What is the proper use of callbacks?

CODE (simplified)

class NODE {
  //...some other stuff
  function create($tags, $callback=false) {
    $temp = new NODE();
    //...code and stuff
    if($callback) $callback($temp); //fixed (from !$callback)
    return $this;
  }
}
$document = new NODE();
$document->create("<p>", function($parent) {
$parent->create("<i>");
});

ERROR

Fatal error: Function name must be a string in P:\htdocs\projects\nif\nif.php on line 36

Can you post your line 36 here. so we can know what is there that's causing the problem? — Miro Markaravanes, May 31 '13 at 22:58
`if(!$callback)` => you only attempt to call the callback when it's `false` (and hence no valid callback...) For safety, you might want to `if(is_callable($callback)) $callback($temp);` — Wrikken, May 31 '13 at 22:59
You can't have such a thing in php. $callback($this) is wrong in syntax if it's false. you should use function pointers if you really want to do this! — Miro Markaravanes, May 31 '13 at 23:01
@MiroMarkarian so there is no way to JS-like callbacks in PHP? — Are, May 31 '13 at 23:08
@AreWojciechowski: Not if your callbacks are `=== false` due to you explicitely preventing anything else from going through an if. Also, `Callable` was added in 5.1 and formally put in 5.3, so if you're running less than that, they won't work. — Sébastien Renauld, May 31 '13 at 23:11
@AreWojciechowski Yes there is a way actually. See my answer. — Miro Markaravanes, May 31 '13 at 23:16
People. If your answer is not an answer to the question, it **goes in the comments**. — Sébastien Renauld, May 31 '13 at 23:19

score 3 · Accepted Answer · answered May 31 '13 at 22:58

3

$document->new NODE();

This is not valid syntax. The accepted format would be:

$document = new NODE();

In addition to this, if you use the unary operator (!) on a false, you get true. If you use it on a Callable, you get false. As such, if (!$callback) $callback() will throw the first error of your script.

As a side note, you are reinventing the wheel. I would strongly recommend you take a look at the DOMDocument family of classes, which are doing exactly what you are currently trying to implement, albeit with fewer callbacks.

answered May 31 '13 at 22:58

Sébastien Renauld

19,203
2
46
66

oops, it should be `document = new NODE();`, fixing. – Are May 31 '13 at 23:00
Yes, I know `DOMDocument` but i need to write my own library for few purposes. – Are May 31 '13 at 23:06
1

@AreWojciechowski: What is your purpose? You are aware that you can extend `DOMDocument`, right? If that doesn't fit the bill, you can also write a `Mediator` or `Adaptor` for it to do what you want to do. – Sébastien Renauld May 31 '13 at 23:07
I can't use external libraries or rather - I don't want to. – Are May 31 '13 at 23:09
1

@AreWojciechowski: Any better reason than "don't want to"? – Sébastien Renauld May 31 '13 at 23:11
+1 (and more if I could give it) for just `extend`ing DOMDocument. – Wrikken May 31 '13 at 23:11
'external'? [You have to explicitly disable it during compile time to remove it from PHP](http://www.php.net/manual/en/dom.installation.php). – Wrikken May 31 '13 at 23:14
@Wrikken by external I meant "written by me". it's for exercise. – Are May 31 '13 at 23:17
Ah well, just make sure your exercise doesn't end up in production is my thought on the matter. At any rate, the comment about the `!` not supposed to be there holds true. – Wrikken May 31 '13 at 23:18

Antoine · Answer 2 · 2013-05-31T23:07:47.860

1

if(!$callback) $callback($temp);

should probably be

if($callback) $callback($temp);

And the instanciation:

$document = new NODE();

edited May 31 '13 at 23:07

answered May 31 '13 at 23:00

Antoine

2,785
1
16
23

You are right about him having the `if` backwards, you are _wrong_ about having to provide a string rather then a function. – Wrikken May 31 '13 at 23:02
1

Sorry, had the call_user_func function in mind when doing this. Thanks for pointing this out – Antoine May 31 '13 at 23:07

score 1 · Answer 3 · answered May 31 '13 at 23:00

1

if(!$callback) $callback($temp);

If $callback is false, for sure you won't be able to call it as a callback.

answered May 31 '13 at 23:00

moonwave99

21,957
3
43
64

`function create($tags, $callback=false) { ... }` $callback is false by default, if not defined by user. if is anything else - executes. right? (i'm not sure if it's good) – Are May 31 '13 at 23:02

score 0 · Answer 4 · answered May 31 '13 at 23:17

0

My 2c here, type hinting may be good to use here as well.

Ex: function create($tags, callable $callback = function())

answered May 31 '13 at 23:17

Rob W

9,134
1
30
50

score -1 · Answer 5 · edited May 23 '17 at 12:18

-1

To do such a thing in php you should use function pointers and tell php which function to execute. Look at this code.

// This function uses a callback function.
function doIt($callback)
{
    $data = acquireData();
    $callback($data);
}


// This is a sample callback function for doIt().
function myCallback($data)
{
    echo 'Data is: ', $data, "\n";
}


// Call doIt() and pass our sample callback function's name.
doIt('myCallback');

So as you seen you can only pass the name to the function and you should predefine the function..

Similar question: How do I implement a callback in PHP?

edited May 23 '17 at 12:18

Community

1
1

answered May 31 '13 at 23:15

Miro Markaravanes

3,285
25
32

Completely off-topic and dangerous malpractice. By doing this, you are polluting the global namespace and forcing PHP to do a hash lookup on explicit name, something you would not have to do otherwise (passing a `Callable` would be one of them). – Sébastien Renauld May 31 '13 at 23:18
@SébastienRenauld I'm just trying to give the OP what he wants. and this is not off-topic. see the comments in the OP and you will see why I posted this answer. and stop down-voting to get your answer up. – Miro Markaravanes May 31 '13 at 23:22
His question was about the callback not working, not about creating a **doIt** function. His way of doing it is miles better than yours - he passes a function pointer/`Callable`, you pass a string reference (= significantly more memory, causes two PHP internal lookups, causes a below-the-bonnet `call_user_func` call). ***THIS*** is why you got a downvote - I don't have a care in the world about my answer. What I care about is not having OT (and blatantly incorrect) answers on a question. – Sébastien Renauld May 31 '13 at 23:25
And just to add water to the mill, this "method" which you copy-pasted from another question dates back from 12/08, when PHP **4** was the de-facto standard. That version of PHP did not have `Callable`. It's 5 years in the future now, and function pointers can be passed to functions. – Sébastien Renauld May 31 '13 at 23:29
@SébastienRenauld mate calm down. I didn't say I wrote the code. Did I? There is no need to "add water to the mill". I just posted my answer. either the best solution or the worst, this is my answer. You could simply comment saying this is not a good way to do it. instead of wowing about this being OT when it's actually A WORKING SOLUTION. – Miro Markaravanes May 31 '13 at 23:33
Miro: I said exactly that in my first comment. You're the one who went personal, "mate". I (rightfully) called your method a dangerous malpractice and explained exactly why it was. You immediately went *ad hominem*. – Sébastien Renauld May 31 '13 at 23:34

Callback in function (PHP) is not working

CODE (simplified)

ERROR

5 Answers5