如何使用Guzzle6发出此POST请求

My working Guzzle5 code looks roughly as follows:

$guzzle = new \GuzzleHttp\Client();
$request = $guzzle->createRequest('POST', $url);
$request->setHeader('Authorization', 'Bearer ' . $token);
$postBody = $request->getBody();
$postBody->setField('name', 'content');//several times
if (check for file) {
    $postBody->addFile(new \GuzzleHttp\Post\PostFile('name', fopen(...));
}
$response = $guzzle->send($request);

What with setting a header and maybe adding a file, I’m not sure how to do this with Guzzle6.

Here an example from the official documentation how can you set headers and adding file into your POST request with Guzzle 6:

$client = new \GuzzleHttp\Client();
$client->post('/post', [
    'multipart' => [
        [
            'name'     => 'foo',
            'contents' => 'data',
            'headers'  => ['X-Baz' => 'bar']
        ],
        [
            'name'     => 'baz',
            'contents' => fopen('/path/to/file', 'r')
        ],
        [
            'name'     => 'qux',
            'contents' => fopen('/path/to/file', 'r'),
            'filename' => 'custom_filename.txt'
        ],
    ]
]);

The multipart option sets the body of the request to a multipart/form-data form, if you don't need to work with files you can just use form_params instead of multipart option.

Any headers you can easy set with help headers option.

Additional info you can find here Guzzle Upgrade Guide (5.0 to 6.0)

Here is some code copied from one of my projects:

$client = new GuzzleHttp\Client();
$url = 'someurl.com/api';
$body = json_encode([
    'variable1' => 'this',
    'variable2' => 'that'
]);


$response = $client->post($url, [
    'headers' => [
        'header_variable1' => 'foo',
        'header_variable2' => 'bar'
    ],
    'json' => true,
    'timeout' => 3,
    'body' => $body
]);

$data = $response->json();