获取在PHP中调用函数的类的名称

Question

I have two separate classes where one of them is a logging class. I would like to be able to tell which class is calling the log class functions without passing any parameters.

<?php
class Log {
    public function general($message) {
        // Tell which class/function is calling this function here
        $class = get_called_class();

        echo 'Your message was: "'.$message.'" from class: "'.$class.'"';
    }
}

class foo {
    public function log_something() {
        $Log = new Log();
        $Log->general('Hello, world!');
    }
}

$foo = new Foo();
$foo->log_something();

?>

The output I would like is: Your message was "Hello, world!" from class "foo"

However, the message I'm getting is: Your message was "Hello, world!" from class "Log"

Am I missing something, or doing something wrong?

Answer 1

I was able to figure out the answer to my problem by using debug_backtrace()

<?php
class Log {
    private function get_call_class() {
        $backtrace = debug_backtrace();

        $classes = array();
        foreach ($backtrace as $item) {
            if ($item['class'] != '' && $item['class'] != 'Log') $classes[] = $item['class'];
        }
        return $classes[0];
    }

    public function general($message) {
        // Tell which class/function is calling this function here
        $class = $this->get_call_class();

        echo 'Your message was: "'.$message.'" from class: "'.$class.'"';
    }
}


class foo {
    public function log_something() {
        $Log = new Log();
        $Log->general('Hello, world!');
    }
}

$foo = new Foo();
$foo->log_something();

?>

Explanation: The Log->get_call_class() function gets information from debug_backtrace() and loops through it to find the classes that don't have the name of 'Log' or blank names then returns the first result, which is the one I was looking for.