对于《Guzzle 5》，《Guzzle 6》和《Guzzle 7》，你是这样做的:

use GuzzleHttp\Client;

$client = new Client();

$response = $client->post('url', [
    GuzzleHttp\RequestOptions::JSON => ['foo' => 'bar'] // or 'json' => [...]
]);

Docs

上述答案对我来说并不管用。但这对我来说很好。

 $client = new Client('' . $appUrl['scheme'] . '://' . $appUrl['host'] . '' . $appUrl['path']);

 $request = $client->post($base_url, array('content-type' => 'application/json'), json_encode($appUrl['query']));

对于Guzzle <= 4:

这是一个原始的post请求，所以把JSON放在body中解决了这个问题

$request = $this->client->post(
    $url,
    [
        'content-type' => 'application/json'
    ],
);
$request->setBody($data); #set body!
$response = $request->send();

$client = new \GuzzleHttp\Client(['base_uri' => 'http://example.com/api']);

$response = $client->post('/save', [
    'json' => [
        'name' => 'John Doe'
    ]
]);

return $response->getBody();

Php版本:5.6

Symfony版本:2.3

暴食:5.0

我最近有一次用Guzzle发送json的经历。我使用的是Symfony 2.3，所以我的暴饮暴食版本可以稍微老一点。

我还会展示如何使用调试模式，你可以在发送请求之前看到它，

当我提出如下所示的请求时，得到了成功的响应;

use GuzzleHttp\Client;

$headers = [
        'Authorization' => 'Bearer ' . $token,        
        'Accept'        => 'application/json',
        "Content-Type"  => "application/json"
    ];        

    $body = json_encode($requestBody);

    $client = new Client();    

    $client->setDefaultOption('headers', $headers);
    $client->setDefaultOption('verify', false);
    $client->setDefaultOption('debug', true);

    $response = $client->post($endPoint, array('body'=> $body));

    dump($response->getBody()->getContents());

$client = new \GuzzleHttp\Client();

$body['grant_type'] = "client_credentials";
$body['client_id'] = $this->client_id;
$body['client_secret'] = $this->client_secret;

$res = $client->post($url, [ 'body' => json_encode($body) ]);

$code = $res->getStatusCode();
$result = $res->json();