guzzle ver 6 post方法不是woking

Question

正在使用guzzle6在邮递员(带有application/json类型的原始格式数据)中工作

url-http://vm.xxxxx.com/v1/hirejob/
{
        "company_name":" company_name",
        "last_date_apply":"06/12/2015",
        "rid":"89498"
}

所以我得到了响应201创建,
但在guzzle

    $client = new Client();
    $data = array();
    $data['company_name'] = "company_name";
    $data['last_date_apply'] = "06/12/2015";
    $data['rid'] = "89498";
    $url='http://vm.xxxxx.com/v1/hirejob/';
    $data=json_encode($data);
    try {
            $request = $client->post($url,array(
                    'content-type' => 'application/json'
            ),array());

        } catch (ServerException $e) {
          //getting GuzzleHttp\Exception\ServerException Server error: 500
        }

我收到了错误 vendor/guzzlehttp/guzzle/src/Middleware.php

第69行

 ? new ServerException("Server error: $code", $request, $response)

Answer 1

你实际上并没有设置请求体,但可以说最简单的传递JSON数据的方法是使用专用的请求选项:

$request = $client->post($url, [
    'json' => [
        'company_name' => 'company_name',
        'last_date_apply' => '06/12/2015',
        'rid' => '89498',
    ],
]);