flock() 是否仅适用于当前方法？

Question

flock() 函数是否仅在与执行代码相同的方法中使用时才有效？

比如下面的代码，加锁成功：

public function run()
{
    $filePointerResource = fopen('/tmp/lock.txt', 'w');
    if (flock($filePointerResource, LOCK_EX)) {
        sleep(10);
    } else {
        exit('Could not get lock!');
    }
}

但是，在下面的代码中，锁是不成功的：

public function run()
{
    if ($this->lockFile()) {
        sleep(10);
    } else {
        exit('Could not get lock!');
    }
}

private function lockFile()
{
    $filePointerResource = fopen('/tmp/lock.txt', 'w');
    return flock($filePointerResource, LOCK_EX);
}

我没有看到任何关于此的文档，所以我对这种行为感到困惑。我使用的是 php 版本 5.5.35。

Answer 1

我认为基于类的尝试的问题在于，当 lockFile 方法完成时，$filePointerResource 超出范围，这可能就是释放锁的原因

这在支持该理论的情况下起作用

<?php

class test {
    public function run()
    {
        $fp = fopen('lock.txt', 'w');
        if ($this->lockFile($fp)) {
            echo 'got a lock'.PHP_EOL;
            sleep(5);
        } 
        /*
         * Not going to do anything as the attempt to lock EX will
         * block until a lock can be gained 

        else {
            exit('Could not get lock!'.PHP_EOL);
        }
        */
    }

    private function lockFile($fp)
    {
        return flock($fp, LOCK_EX);
    }
}

$t = new test();
$t->run();

因此，如果您想在多次调用类方法时锁定文件，最好将文件句柄保留为类属性，那么只要类被实例化并在范围内，它就会保持在范围内。

<?php

class test {
    private $fp;

    public function run()
    {
        $this->fp = fopen('lock.txt', 'w');
        if ($this->lockFile()) {
            echo 'got a lock'.PHP_EOL;
            sleep(5);
        } 
        /*
         * Not going to do anything as the attempt to lock EX will
         * block until a lock can be gained 

        else {
            exit('Could not get lock!'.PHP_EOL);
        }
        */
    }

    private function lockFile()
    {
        return flock($this->fp, LOCK_EX);
    }
}

$t = new test();
$t->run();