Question

我有两个单独的类，其中一个是日志记录类。我希望能够告诉哪个类在不传递任何参数的情况下调用日志类函数。

<?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();

?>

我想要的输出是：Your message was "Hello, world!" from class "foo"

然而，我得到的信息是： Your message was "Hello, world!" from class "Log"

我错过了什么，或做错了什么？

Answer 1

您只需在Log内创建Foo对象，因此get_called_class()来电将始终生成Log。如果您想要这种行为，则需要Foo继承Log。例如：

<?php

class Log {
    public function msg($message) {
        // Tell which class/function is calling this function here
        $class = get_called_class();

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

class Foo extends Log {
  public function doSomething() {
    $this->msg( 'Hello World!' );
  }
}

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

输出：

Your message was: "Hello World!" from class: "Foo"

Answer 2

我能够使用debug_backtrace()

找出问题的答案

<?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();

?>

说明：Log->get_call_class()函数从debug_backtrace()获取信息并循环显示该信息，以查找不具有＆＃39; Log＆＃39;或者空白名称然后返回第一个结果，这是我正在寻找的结果。

获取在PHP中调用函数的类的名称

2 个答案: