如何发布Guzzle 6异步数据

Question

我尝试使用Guzzle 6(最新版本)以异步方式发布数据

    $client = new Client();
    $request = $client->postAsync($url, [
        'json' => [
                'company_name' => 'update Name'
        ],
    ]);

但是我在终端上没有收到任何类似于post请求的请求

Answer 1

因为它是一个promise,，所以你需要把then

并且promise不会调用，除非您将$promise->wait()

根据您的问题，这是一个使用postAsync的简单post请求：

$client = new Client();
$promise = $client->postAsync($url, [
    'json' => [
            'company_name' => 'update Name'
    ],
])->then(
    function (ResponseInterface $res){
        $response = json_decode($res->getBody()->getContents());

        return $response;
    },
    function (RequestException $e) {
        $response = [];
        $response->data = $e->getMessage();

        return $response;
    }
);
$response = $promise->wait();
echo json_encode($response);

Answer 2

您是否已尝试发送请求？

http://guzzle.readthedocs.org/en/latest/index.html?highlight=async

$client = new Client();
$request = new Request('POST', $url, [
    "json" => [
        'company_name' => 'update Name']
    ]);

$promise = $client->sendAsync($request)->then(function ($response) {
    echo 'I completed! ' . $response->getBody();
});
$promise->wait();

Answer 3

您最有可能需要调用wait();

$request->wait();