PHP GuzzleHttp。如何使用参数发出 post 请求？

Question

如何使用 GuzzleHttp（版本 5.0）发出 post 请求。

我正在尝试执行以下操作：

$client = new \GuzzleHttp\Client();
$client->post(
    'http://www.example.com/user/create',
    array(
        'email' => 'test@gmail.com',
        'name' => 'Test user',
        'password' => 'testpassword'
    )
);

但我收到错误消息：

PHP Fatal error: Uncaught exception 'InvalidArgumentException' with the message 'No method can handle the email config key'

Answer 1

试试这个

$client = new \GuzzleHttp\Client();
$client->post(
    'http://www.example.com/user/create',
    array(
        'form_params' => array(
            'email' => 'test@gmail.com',
            'name' => 'Test user',
            'password' => 'testpassword'
        )
    )
);

Answer 2

由于 Marco 的回答已弃用，您必须使用以下语法（根据 jasonlfunk 的评论）：

$client = new \GuzzleHttp\Client();
$response = $client->request('POST', 'http://www.example.com/user/create', [
    'form_params' => [
        'email' => 'test@gmail.com',
        'name' => 'Test user',
        'password' => 'testpassword',
    ]
]);

请求 POST 个文件

$response = $client->request('POST', 'http://www.example.com/files/post', [
    'multipart' => [
        [
            'name'     => 'file_name',
            'contents' => fopen('/path/to/file', 'r')
        ],
        [
            'name'     => 'csv_header',
            'contents' => 'First Name, Last Name, Username',
            'filename' => 'csv_header.csv'
        ]
    ]
]);

REST 动词与参数一起使用

// PUT
$client->put('http://www.example.com/user/4', [
    'body' => [
        'email' => 'test@gmail.com',
        'name' => 'Test user',
        'password' => 'testpassword',
    ],
    'timeout' => 5
]);

// DELETE
$client->delete('http://www.example.com/user');

异步POST数据

对于长时间的服务器操作很有用。

$client = new \GuzzleHttp\Client();
$promise = $client->requestAsync('POST', 'http://www.example.com/user/create', [
    'form_params' => [
        'email' => 'test@gmail.com',
        'name' => 'Test user',
        'password' => 'testpassword',
    ]
]);
$promise->then(
    function (ResponseInterface $res) {
        echo $res->getStatusCode() . "\n";
    },
    function (RequestException $e) {
        echo $e->getMessage() . "\n";
        echo $e->getRequest()->getMethod();
    }
);

设置headers

根据documentation，可以设置headers：

// Set various headers on a request
$client->request('GET', '/get', [
    'headers' => [
        'User-Agent' => 'testing/1.0',
        'Accept'     => 'application/json',
        'X-Foo'      => ['Bar', 'Baz']
    ]
]);

更多调试信息

如果您需要更多详细信息，可以使用 debug 选项，如下所示：

$client = new \GuzzleHttp\Client();
$response = $client->request('POST', 'http://www.example.com/user/create', [
    'form_params' => [
        'email' => 'test@gmail.com',
        'name' => 'Test user',
        'password' => 'testpassword',
    ],
    // If you want more informations during request
    'debug' => true
]);

Documentation 更明确地说明了新的可能性。

Answer 3

请注意，在 Guzzle V6.0+ 中，出现以下错误的另一个原因可能是错误地将 JSON 用作数组：

Passing in the "body" request option as an array to send a POST request has been deprecated. Please use the "form_params" request option to send a application/x-www-form-urlencoded request, or a the "multipart" request option to send a multipart/form-data request.

不正确:

$response = $client->post('http://example.com/api', [
    'body' => [
        'name' => 'Example name',
    ]
])

正确:

$response = $client->post('http://example.com/api', [
    'json' => [
        'name' => 'Example name',
    ]
])

正确:

$response = $client->post('http://example.com/api', [
    'headers' => ['Content-Type' => 'application/json'],
    'body' => json_encode([
        'name' => 'Example name',
    ])
])

Answer 4

$client = new \GuzzleHttp\Client();
$request = $client->post('http://demo.website.com/api', [
    'body' => json_encode($dataArray)
]);
$response = $request->getBody();

添加

openssl.cafile 在 php.ini 文件中

PHP GuzzleHttp。如何使用参数发出 post 请求？

PHP GuzzleHttp. How to make a post request with params?

php

httpclient

request

guzzle

请求 POST 个文件

REST 动词与参数一起使用

异步POST数据

设置headers

更多调试信息