Question

有人要求在此问题中注册变量更改的事件处理程序：PHP how to detect the change of variable?

我尝试使用PHP的魔术函数__get和__set开发一个快速类。这有效，直到我通过引用将成员传递给普通函数，它不再触发事件。

这是一个错误，还是一些不可能的错误，或者我只是错过了什么？

<?php
header("content-type: text/plain");

class WatchVar {
    private $data = array();
    private $org = array();
    private $callbacks = array();

    public function __set($name, $value) {
        if (!array_key_exists($name, $this->data)) {
            $this->org[$name] = $value;
        } else {
            //variable gets changed again!
            $this->triggerChangedEvent($name, $value);
        }
        $this->data[$name] = $value;
    }

    public function &__get($name) {
        if (array_key_exists($name, $this->data)) {
            if ($this->data[$name] != $this->org[$name]) {
                //variable has changed, return original
                //return $this->org[$name];
                //or return new state:
                return $this->data[$name];
            } else {
                //variable has not changed
                return $this->data[$name];
            }
        }
    }

    public function addCallback($name, $lambdaFunc) {
        $this->callbacks[$name] = $lambdaFunc;
    }

    protected function triggerChangedEvent($name, $value) {
        //$this->data[$name] has been changed!
        //callback call like:
        call_user_func($this->callbacks[$name], $value);
    }
}


$test = new WatchVar;
$test->addCallback('xxx', function($newValue) { echo "xxx has changed to {$newValue}\n"; });
$test->xxx = "aaa";

echo $test->xxx . "\n";
//output: aaa

$test->xxx = "bbb";
//output: xxx has changed to bbb

echo $test->xxx . "\n";
//output bbb

function messyFunction(&$var) {
    $var = "test";
}

messyFunction($test->xxx);
//output: nothing, why?

Answer 1

改变这段代码：

function messyFunction(&$var) {
    $var->xxx = "test";
}

messyFunction($test);
//output: xxx has changed to test

//输出：没有，为什么？

即使通过引用传递，该函数也只接收成员变量的克隆而不是实例+魔术函数。

PHP：通过引用传递类成员不起作用

1 个答案: