在lumen api中从Guzzle post请求中获取form_params数据

Question

public function insert(Request $request)
    { 
        $domain = Config::get('url.gcr_api'); // gcr_api is defined in app/config/url.php
        $url = $domain . "/api/v1/campaign/insert"; 

        $campaign_name =  $request->input('campaign_name');
        $campaign_provider =  $request->input('campaign_provider');

        $client = new Client;
        $response = $client->request('POST', $url , [
                                'headers' => [
                                    'X-XSRF-TOKEN' => csrf_token(),
                                ],
                                'form_params' => [
                                    'name' => $campaign_name, 
                                    'bank_id' => $campaign_provider,
                                ]
                            ]
                    );

        $data = json_decode($response->getBody(), true);
        return $data;

    }
}

下面是上述方法调用的lumen api的代码。

public function insert(Request $request) {
        try {
            $campaign_name =  $request->input('campaign_name');
            $campaign_provider =  $request->input('campaign_provider');

            $data = array('name' => $campaign_name, 'bank_id' => $campaign_provider);
            DB::table('campaign')->insert($data); 

            $response = array(
                'status' => 'success',
                'message' => 'Campaign created successfully',
            );        
            return response()->json($response);
        }
        catch(Exception $e) {
            // When query fails. 
            $response = array(
                'status' => 'failed',
                'message' => 'Failed to create campaign',
            );

            return response()->json($response);
        }
    }

每次调用laravel应用程序控制器的insert方法时，我都会得到这个错误：

GuzzleHttp\Exception\ServerException(500)服务器错误:发送http://localhost:8000/api/v1/campaign/insert导致500内部服务器错误响应：

/*$campaign_name =  $request->input('campaign_name');
  $campaign_provider =  $request->input('campaign_provider');

  $data = array('name' => $campaign_name, 'bank_id' => $campaign_provider);
  DB::table('campaign')->insert($data); */

Answer 1

您的假设部分正确，它可以访问form_params，但是您试图获取的form_params与实际参数不同：

为了传递paramters，可以使用以下数组：

'form_params' => [
    'name' => $campaign_name, 
    'bank_id' => $campaign_provider,
]

此数组将值设置为$campaign_name，键为'name'，并将$campaign_provider设置为bank_id。到目前为止，这段代码没有问题。

$campaign_name =  $request->input('campaign_name');
$campaign_provider =  $request->input('campaign_provider');

$campaign_name =  $request->input('name');
$campaign_provider =  $request->input('bank_id');