调用前检查某项是否为函数

Question

I have a class that looks a bit like this:

<?php

namespace App\Http\Controllers;

use Exception;
use Illuminate\Http\Request;

class FormatAddressController extends Controller
{
    public function __construct()
    {
        $this->middleware(['api-auth']);
        $this->middleware(['api-after']);
    }

    public function format(Request $request) {
        // TODO first, get customer settings to see what functions to run, and how to run them
        // but, assume that the settings come as an array where the key is the function name
        // and the value is one of NULL, false, or settings to pass through to the function
        $settings = ['isoAndCountry' => true, 'testFunc' => ['testSetting' => 'test setting value']];

        $address = $request->input('address');

        $errors = [];

        foreach ($settings as $funcName => $funcSettings) {    
            try {
                $address = $this->$funcName($funcSettings, $address); // every function has to return the modified address
            } catch(Exception $e) {
                $errors[$funcName] = $e;
            }

        }

        return response()->json([
            'address' => $address,
            'errors' => $errors
        ]);
    }

    public function isoAndCountry($settings, $address) {
        // TODO

        return $address;
    }
}

Now, when I call this function, isoAndCountry, through that settings loop I defined above, it works! It works just fine.

However I tried following this thread and checking is_callable and... it errors:

            if (is_callable($this->$funcName)) {
                try {
                    $address = $this->$funcName($funcSettings, $address); // every function has to return the modified address
                } catch(Exception $e) {
                    $errors[$funcName] = $e;
                }
            }

How can I check if it's callable? Why doesn't this work?

Answer 1

可能这也可以解决问题：

 if(method_exists($this,$funcName)){ ... }

Answer 2

You have to use method_exists here to check if the method really exists in the class.

foreach ($settings as $funcName => $funcSettings) {
    if (method_exists($this, $funcName)) {
        $this->$funcName($funcSettings, $address);
    }
}

The reason why is_callable will not work in your scenario is because Laravel controllers has a __call magic method which will handle undefined methods, so running is_callable on any non existing methods would return true.

Take the following class as an example:

class A
{
    public function __construct()
    {
        var_dump(is_callable([$this, 'testFunc']));
    }
}

The output of new A would be false . However, if you add the following into the class:

public function __call($name, $arguments)
{
    //
}

Now the output of the var_dump would return true .

You can read more about the __call scenario I've mentioned right here: https://www.php.net/manual/en/function.is-callable.php#118623

For more information about __call : https://www.php.net/manual/en/language.oop5.overloading.php#object.call

Answer 3

You can use

if (is_callable([$this, $funcName])) { ...

instead.

The way you have it written with is_callable($this->$funcName) , it's going to look for a property called $funcName on $this , (which probably doesn't exist) and check if that's callable. If you use that array syntax it will evaluate the named method instead.

---

It may be simpler in this case to use

if (method_exists($this, $funcName)) {

since you're using it in another method of the same object, if the method exists it should be callable.

Answer 4

You need to differentiate between class properties and methods. For example in this class:

class A {
    private $foo = null;

    public function getFoo() {
        return $this->foo;
    }
}

• the private $foo is property and can be checked by property_exists()
• the public function getFoo() is method and can be checked by method_exists()

https://www.php.net/manual/en/function.property-exists.php
https://www.php.net/manual/en/function.method-exists.php

I think that $this->$funcName works only for properties.