flock()函数是否仅在执行代码的同一方法中使用时才起作用？
例如，在以下代码中，锁定成功：
public function run()
{
$filePointerResource = fopen('/tmp/lock.txt', 'w');
if (flock($filePointerResource, LOCK_EX)) {
sleep(10);
} else {
exit('Could not get lock!');
}
}
但是，在以下代码中，锁定不成功：
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);
}
我还没有看到任何关于这方面的文件，所以我对这种行为感到困惑。我使用的是php版本5.5.35。
我认为基于类的尝试的问题是，当lockFile方法完成时，$filePointerResource会超出范围，这可能就是释放锁定的原因
这在某种程度上支持理论
<?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，如果您想通过对类方法的多个调用来锁定文件，最好将filehandle保留为类属性，那么只要类被实例化并处于作用域中，它就会保持在作用域中。
<?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();