Laravel自定義驗證 - Mailgun API

Question

我想實現的Laravel的新功能「自定義驗證規則」之一，我遇到了以下錯誤：

Object of class Illuminate\Validation\Validator could not be converted to string 

我按照此步驟視頻： New in Laravel 5.5: Project: Custom validation rule classes (10/14)

這是一個嘗試Mailgun API的電子郵件驗證工具。

請求簡單的形式：姓，名，公司，電子郵件和消息

這裏是我的代碼：

web.php

Route::post('contact', 'StaticPageController@postContact'); 

StaticPageController.php

use Validator; 
use App\Http\Validation\ValidEmail as ValidEmail; 

public function postContact(Request $request) { 
     return Validator::make($request->all(), [ 
      'firstname' => 'required|max:90', 
      'lastname' => 'required|max:120', 
      'company' => 'max:120', 
      'email' => [ 
       'required', 'string', 'max:255', 
       new ValidEmail(new \GuzzleHttp\Client) 
      ], 
      'message' => 'required', 
     ]); 
} 

ValidEmail.php

<?php 

namespace App\Http\Validation; 

use Illuminate\Contracts\Validation\Rule; 
use GuzzleHttp\Exception\GuzzleException; 
use GuzzleHttp\Client as Guzzle; 

class ValidEmail implements Rule 
{ 
    protected $client; 
    protected $message = 'Sorry, invalid email address.'; 

    public function __construct(Guzzle $client) 
    { 
     $this->client = $client; 
    } 

    public function passes($attribute, $value) 
    { 
     $response = $this->getMailgunResponse($value); 
    } 

    public function message() 
    { 
     return $this->message; 
    } 

    protected function getMailgunResponse($address) 
    { 
     $request = $this->client->request('GET', 'https://api.mailgun.net/v3/address/validate', [ 
      'query' => [ 
       'api_key' => env('MAILGUN_KEY'), 
       'address' => $address 
      ] 
     ]); 
     dd(json_decode($request->getBody())); 
    } 
} 

期望

我期待看到這樣的事情：

{ 
    +"address": "test@e2.com" 
    +"did_you_mean": null 
    +"is_disposable_address": false 
    +"is_role_address": false 
    +"is_valid": false 
    +"parts": { 
     ... 
    } 
} 

任何幫助深表感謝。我一直試圖讓這個簡單的例子現在工作兩個多小時。希望有我的經驗的人可以幫忙！

Answer 1

在你的控制器

試試這個：

$validator = Validator::make($request->all(), [ 
    'firstname' => 'required|max:90', 
    'lastname' => 'required|max:120', 
    'company' => 'max:120', 
    'email' => [ 
     'required', 'string', 'max:255', 
     new ValidEmail(new \GuzzleHttp\Client) 
    ], 
    'message' => 'required', 
]); 


if ($validator->fails()) { 
    return redirect()->back() 
     ->withErrors($validator) 
     ->withInput(); 
} 

// if valid ... 

Answer 2

根據您的路線，該postContact方法是處理路由的方法。這意味着此方法的返回值應該是您想要查看的響應。

您正在返回一個Validator對象，然後Laravel試圖將其轉換爲響應的字符串。 Validator對象不能轉換爲字符串。

您需要執行驗證，然後根據驗證返回正確的響應。您可以在documenation here中閱讀更多關於手動驗證器的信息。

總之，你需要的是這樣的：

public function postContact(Request $request) { 
    $validator = Validator::make($request->all(), [ 
     'firstname' => 'required|max:90', 
     'lastname' => 'required|max:120', 
     'company' => 'max:120', 
     'email' => [ 
      'required', 'string', 'max:255', 
      new ValidEmail(new \GuzzleHttp\Client) 
     ], 
     'message' => 'required', 
    ]); 

    // do your validation 
    if ($validator->fails()) { 
     // return your response for failed validation 
    } 

    // return your response on successful validation 
}