Question

我正在尝试使用Guzzle在我的laravel应用程序中使用第三方API，但是我收到了400 Bad Request错误。以下是控制器的代码

<?php
public function sendEmailAmazon(Request $request)
{
    $data = json_decode(json_encode($request->all()));
    $mailaddresses = $data->tomail;

    // dd($mailaddresses);

    foreach ($mailaddresses as $mailaddress) {
        $client = new Client();
        $value = Config::get('app');
        $maildata = array(
            'mailId' => null,
            'type' => '1',
            'from' => array(
                'name' => $value['from_name'],
                'address' => $value['from_email'],
                'mobile' => $value['from_mobile'],
            ),
            'to' => $data->tomail,
            'cc' => null,
            'replyTo' => null,
            'subject' => $data->subject,
            'body' => $data->msg,
            'text' => null,
            'attachment' => null,
            'personalization' => array(
                'salutation' => 'Dear',
                'useFirstNameOnly' => true
            ),
            'apiKey' => $value['api_key'],
            'apiSecret' => $value['api_secret'],
            'customerId' => $value['customer_id'],
        );
        $data2 = json_encode($maildata);
        $url = $value['api_url'];
        try {
            $res = $client->post($url, ['Accept' => 'application/json', 'json' => $data2, ]);
            return $res;
            $this->addEmailHistory($data);
            return response()->json(['success' => 'true', 'data' => $data2, 'message' => 'Email sent']);
        } catch(Exception $ex) {
            return Common::getJsonResponse(false, $ex->getMessage() , 300);
        }
    }
}

我还尝试了documentation中所述的代码，但仍然遇到同样的错误：

$r = $client->request('POST', $url, [
    'json' => $data2
]);

我的路线

Route::post('email/sendEmailAmazon', 'EmailsController@sendEmailAmazon');

Answer 1

Your error may be from api server. If your api also written using laravel, i think this is CSRF token mismatch exception error.

在laravel中使用Guzzle获得“400 Bad Request”响应

1 个答案: