PHP Callable Object as Object Member
I have a class Logger
which, among other things has a method Log
.
Log
is the most common use of the Logger
instance, I have wired __invoke
to call Log
Another class, "Site" contains a member "Log", an instance of Logger.
Why would this work:
$L开发者_JAVA百科og = $this->Log;
$Log("Message");
But not this:
$this->Log("Message");
The former fails with "PHP Fatal error: Call to undefined method Site::Log()"
Is this a limitation of the callable object implementation, or am I misunderstanding something?Unfortunately, this is (still) a limitation of PHP, but it makes sense when you think about it, as a class can contain properties and methods that share names. For example:
<?php
class Test {
public $log;
public function __construct() {
$this->log = function() {
echo 'In Test::log property';
};
}
public function log() {
echo 'In Test::log() method';
}
}
$test = new Test;
$test->log(); // In Test::log() method
call_user_func($test->log); // In Test::log property
?>
If PHP were to allow the syntax you desire, which function would be invoked? Unfortunately, that only leaves us with call_user_func[_array]()
(or copying $this->log
to another variable and invoking that).
However, it would be nice if the following syntax was acceptable:
<?php
{$test->log}();
?>
But alas, it is not.
Same reasons you can't do this:
$value = $this->getArray()["key"];
or even this
$value = getArray()["key"];
Because PHP syntax doesn't do short hand very well.
This may work:
${this->Log}("Message");
But perhaps it's just easier and better to use the full call? There doesn't seem to be a way to get what you want to work on the one line.
The error in your question indicates it is looking for a function defined on the class which doesn't exist. An invokable object isn't a function, and it seems it can't be treated as one in this case.
as of php7.4 the following code works for me
($this->Log)("Message");
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