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();

$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();

这适用于我的Guzzle 6.2:

$gClient =  new \GuzzleHttp\Client(['base_uri' => 'www.foo.bar']);
$res = $gClient->post('ws/endpoint',
                            array(
                                'headers'=>array('Content-Type'=>'application/json'),
                                'json'=>array('someData'=>'xxxxx','moreData'=>'zzzzzzz')
                                )
                    );

根据文档guzzle做json_encode

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

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

return $response->getBody();

来自@user3379466的答案可以通过设置$data来工作，如下所示:

$data = "{'some_key' : 'some_value'}";

我们的项目需要的是将一个变量插入到json字符串中的数组中，我这样做如下(如果这有助于任何人):

$data = "{\"collection\" : [$existing_variable]}";

因此，与$existing_variable是，说，90210，你得到:

echo $data;
//{"collection" : [90210]}

另外值得注意的是，你可能还想设置'Accept' => 'application/json'以防你碰到的端点关心这类事情。

简单而基本的方法(guzzle6):

$client = new Client([
    'headers' => [ 'Content-Type' => 'application/json' ]
]);

$response = $client->post('http://api.com/CheckItOutNow',
    ['body' => json_encode(
        [
            'hello' => 'World'
        ]
    )]
);

为了获得响应状态代码和主体的内容，我这样做:

echo '<pre>' . var_export($response->getStatusCode(), true) . '</pre>';
echo '<pre>' . var_export($response->getBody()->getContents(), true) . '</pre>';