如何阻止PHP类执行?

你可以用 PHP exceptions :

// This function would of course be declared in the class
function stop() {
    throw new Exception('Stopped.');
}

try {
    $browser->isStatusCode(200)
      ->isRequestParameter('module', 'home')
      ->isRequestParameter('action', 'index')
      ->click('Register')
      ->stop()
      ->I_DONT_WANT_THIS_TO_RUN();
} catch (Exception $e) {
    // when stop() throws the exception, control will go on from here.
}

$browser->thenThisRunsOkay();

只需返回另一个类,该类将为每个调用的方法返回$this。

例子:

class NoMethods {
    public function __call($name, $args)
    {
        echo __METHOD__ . " called $name with " . count($args) . " arguments.\n";
        return $this;
    }
}

class Browser {
    public function runThis()
    {
        echo __METHOD__ . "\n";
        return $this;
    }

    public function stop()
    {
        echo __METHOD__ . "\n";
        return new NoMethods();
    }

    public function dontRunThis()
    {
        echo __METHOD__ . "\n";
        return $this;
    }
}

$browser = new Browser();
echo "with stop\n";
$browser->runThis()->stop()->dontRunThis()->dunno('hey');
echo "without stop\n";
$browser->runThis()->dontRunThis();
echo "the end\n";

将导致:

with stop
Browser::runThis
Browser::stop
NoMethods::__call called dontRunThis with 0 arguments.
NoMethods::__call called dunno with 1 arguments.
without stop
Browser::runThis
Browser::dontRunThis
the end

OIS的答案非常好,尽管我可以看出,如果对象突然变为其他对象,它可能会变得混乱。也就是说,您希望在链的末尾,您将以相同的对象结束。为了避免这个问题,我添加了一个私有变量来告诉类是否实际执行任何操作。如果类已停止,则每个类都将返回 $this 马上。这为您提供了能够重新启动执行的额外好处。

class MyClass {
    private $halt;

    function __call($func, $args) {
        if ($this->halt) {
            return $this;
        } else {
            return $this->$func($args);
        }
    }

    private function isRequestParameter() {
        // ...
    }
    public function stop() {
        $this->halt = true;
    }
    public function start() {
        $this->halt = false;
    }
}

可以将其放入父类中,这样就不必重复此代码。