Question

我希望在一个循环中执行某些操作，然后在请求结束时继续执行脚本，例如我想做这样的事情：

$client = new \GuzzleHttp\Client();
$promise = $client->headAsync("https://www.google.com");
$promise->then(
  function (ResponseInterface $res) {
    echo $res->getStatusCode() . "\n";
  },
  function (RequestException $e) {
    echo $e->getMessage() . "\n";
  }
);
while($promise->getState() === "pending"){
  $queue = \GuzzleHttp\Promise\queue();
  $queue->run();
  echo "Waiting\n";
  sleep(1);
}

这段代码只打印“等待”。我怎么能实现这个目标呢？

Answer 1

正在寻找相同的答案，所以这是我发现的，你需要tick curl请求（以便请求实际上是进程）。

工作示例：

$curl = new \GuzzleHttp\Handler\CurlMultiHandler();
$handler = \GuzzleHttp\HandlerStack::create($curl);
$client = new \GuzzleHttp\Client(['handler' => $handler]);

$promise = $client->headAsync("https://www.google.com");
$promise->then(
  function (ResponseInterface $res) {
    echo $res->getStatusCode() . "\n";
  },
  function (RequestException $e) {
    echo $e->getMessage() . "\n";
  }
);

$queue = \GuzzleHttp\Promise\queue();
while($promise->getState() === "pending"){
  $curl->tick();
  //echo "Waiting\n"; Commented out as it prints A LOT without sleep, but sleep slows down the rest.
}

在查看此问题后，我将此代码放在一起：https://github.com/guzzle/guzzle/issues/1127

如何在等待期间做其他事情的同时等待guzzle请求完成？

1 个答案: