发送Guzzle的“异步”请求而不调用“等待”

Question

I'm trying to send a request to an endpoint, but I don't want to wait for them to respond, as I don't need the response. So I'm using Guzzle, here's how:

$url = 'http://example.com';

$client = new \Guzzelhttp\Client();
$promise = $client->postAsync($url, [
    'headers' => ['Some headers and authorization'],
    'query' => [
        'params' => 'params',
    ]
])->then(function ($result) {
    // I don't need the result. So I just leave it here.
});

$promise->wait();

A I understood, I have to call the wait method on the client in order to actually send the request. But it's totally negates the request being "async" because if the url was not accessible or the server was down, the application would wait for a timeout or any other errors.

So, the question here is, what does Guzzle mean by "async" when you have to wait for the response anyway? And how can I call a truly async request with PHP?

Thanks

Answer 1

What you can do is:

    $url = 'http://example.com';

    $client = new \Guzzelhttp\Client();
    $promise = $client->postAsync($url, [
        'headers' => ['Some headers and authorization'],
        'query' => [
            'params' => 'params',
        ]
    ])->then(function ($result) {
         return $result->getStatusCode();
    })
->wait();

echo $promise;

You need the wait() to be called as the last line so you get the result which will come from your promise.

In this case it will return just the status code.

Just as mentioned in Github is not able to "fire and forget"so i think what you are trying to achieve, like a complete promise like in Vue or React won't work for you here the way you want it to work.

Another approach and what i do personally is to use a try-catch on guzzle requests, so if there is a guzzle error then you catch it and throw an exception.