GuzzleHTTP在现有页面上返回404

Question

我对https://api.scarif.dev/auth的POST请求返回了404，而该页面通过Postman，浏览器或javascript存在。它应该返回200并显示401消息，但Guzzle会返回404。在POST和GET模式下，都是这样。

我尝试了多种客户端设置，包括不同的标头和禁用SSL验证，但没有成功。现在，我复制了与在邮递员中可用的完全相同的标头，但仍然没有成功。

我一直在搜索google和stackoverflow，但是找不到解决我问题的答案。

PHP请求：

<?php
$client = new Client([
    'header' => [
        'Accept' => 'application/json',
        'Content-Type' => 'application/x-www-form-urlencoded'
    ],
    'verify' => false
]);

$response = $client->request('POST', 'https://api.scarif.dev/auth', [
    'form_params' => []
]);

echo $response->getBody()->getContents();
?>

预期结果：

{
    "detail": "https://login.scarif.dev",
    "status": 401,
    "title": "Unauthorized",
    "type": "http://www.w3.org/Protocols/rfc2616/rfc2616-sec10.html"
}

实际结果：

  

致命错误：未捕获的GuzzleHttp \ Exception \ ClientException：客户端   错误：POST https://api.scarif.dev/auth导致404 Not Found响应：    找不到404

  （被截断...）在   /home/admin/domains/login.scarif.dev/framework/vendor/guzzlehttp/guzzle/src/Exception/RequestException.php:113   堆栈跟踪：＃0   /home/admin/domains/login.scarif.dev/framework/vendor/guzzlehttp/guzzle/src/Middleware.php(66）：   GuzzleHttp \ Exception \ RequestException :: create（Object（GuzzleHttp \ Psr7 \ Request），   对象（GuzzleHttp \ Psr7 \ Response））＃1   /home/admin/domains/login.scarif.dev/framework/vendor/guzzlehttp/promises/src/Promise.php(203）：   GuzzleHttp \ Middleware :: GuzzleHttp {closure}（Object（GuzzleHttp \ Psr7 \ Response））

     2 /home/admin/domains/login.scarif.dev/framework/vendor/guzzlehttp/promises/src/Promise.php(156）：

     

GuzzleHttp \ Promise \ Promise :: callHandler（1，   对象（GuzzleHttp \ Psr7 \ Response），数组）＃3   /home/admin/domains/login.scarif.dev/framework/ven in   /home/admin/domains/login.scarif.dev/framework/vendor/guzzlehttp/guzzle/src/Exception/RequestException.php   在第113行

API端点控制器：

<?php

namespace Controller;

use Core\Config;
use Core\Request;
use Core\Response;
use Model\Token;
use Model\User;
use MongoDB\BSON\UTCDateTime;

class AuthController extends Controller
{
    public function view(User $user, Token $token)
    {
        extract(Request::getPostData());

        if (isset($access_token) && !empty($access_token)) {
            $_token = $token->getTokenByToken($access_token);

            if (
                $_token['type'] !== Token::TYPE_ACCESS_TOKEN ||
                $_token['expires_on'] <= new UTCDateTime()
            ) {
                return $this->view->display('json', [
                    'payload' => Response::apiResponse(
                        $this->config->get('url.login'), 401
                    )
                ]);
            }

            $token->delete($_token['_id']);

            $newToken = $token->create(Token::TYPE_ACCESS_TOKEN, $_token['user_id']);

            return $this->view->display('json', [
                'payload' => Response::apiResponse($newToken['token'])
            ]);
        }

        if (!isset($email) || !isset($password) || empty($email) || empty($password)) {
            return $this->view->display('json', [
                'payload' => Response::apiResponse(
                    $this->config->get('url.login'), 401
                )
            ]);
        }

        if (!$user->checkCredentials($email, $password)) {
            return $this->view->display('json', [
                'payload' => Response::apiResponse(
                    "The email address or password you've entered is invalid. Please check your entry and try again.",
                    422
                )
            ]);
        }

        $user = $user->getUserByEmail($email);
        $token = $token->create(Token::TYPE_ACCESS_TOKEN, $user['_id']);

        return $this->view->display('json', [
            'payload' => Response::apiResponse($token['token'])
        ]);
    }
}

Answer 1

问题似乎出在您使用的API上。将代码与其他网址一起使用时，效果很好：

$client = new Client([
    'header' => [
        'Accept' => 'application/json',
        'Content-Type' => 'application/x-www-form-urlencoded'
    ],
    'verify' => false
]);

$response = $client->request('POST', 'https://jsonplaceholder.typicode.com/posts', [
    'form_params' => []
]);

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

您可以显示API端点的代码吗？