php - Laravel api路由类发现错误

Question

我目前正在使用 SitePoint 上的样板开发基于 api jwt 的身份验证。到目前为止，一切正常，但我仍停留在这一点上。

我的 Controller 看起来像这样:

namespace App\Api\V1\Controllers;

use Illuminate\Http\Request;
use Dingo\Api\Routing\Helpers;

use Symfony\Component\HttpKernel\Exception\HttpException;
use JWTAuth;
use App\Http\Controllers\Controller;
use App\Order;
// use App\Api\V1\Requests\LoginRequest;
use Tymon\JWTAuth\Exceptions\JWTException;
use Symfony\Component\HttpKernel\Exception\AccessDeniedHttpException;

在正文中我有这个功能:

  public function checkThis()
  {
    $currentUser = JWTAuth::parseToken()->authenticate();
    $orders = App\Order::first();
    echo $orders;
    function() {
    echo "stll here";
    };
  }

在我的 api 路由下，我在中间件中有这个:

$api->get('orderlist', 'App\\Api\\V1\\Controllers\\OrderController@checkThis');

当我在 postman 中运行它时，我收到以下错误:“消息”:“找不到类 'App\Api\V1\Controllers\App\Order'”，

我已经尝试了所有我能想到的方法，但它仍然在发生。当我将它从 Controller 中取出并直接在它工作的 route 运行时。我是 Laravel 和 PHP 的新手，所以我有点卡住了。

Answer 1

以下所有内容都表明您要调用 App\Order::first();

在函数 checkThis 中，您可以将 App\Order::first() 替换为

Order::first() //aliasing version

或将 App\Order::first() 替换为

\App\Order::first(); //fully qualified Name version

通过 php manual

Example #1 importing/aliasing with the use operator
<?php
namespace foo;
use My\Full\Classname as Another;

// this is the same as use My\Full\NSname as NSname   <-- very important 
use My\Full\NSname;

注意

// this is the same as use My\Full\NSname as NSname   <-- very important 
    use My\Full\NSname;

another php manual

and Inside a namespace, when PHP encounters an unqualified Name in a class name, function or constant context, it resolves these with different priorities. Class names always resolve to the current namespace name. Thus to access internal or non-namespaced user classes, one must refer to them with their fully qualified Name

php manual : fully qualified Name

Fully qualified name

This is an identifier with a namespace separator that begins with a namespace separator, such as \Foo\Bar. The namespace \Foo is also a fully qualified name.

所以如果你想调用函数App\Order::first，只需Order::first，因为

use App\Order;     
equal 
use App\Order as Order ;

别名是 Order 而不是 App\Order 。 完全限定名称是\App\Order而不是App\Order。

另一方面，当您调用 App\Order::first();这意味着你正在调用

App\Api\V1\Controllers\App\Order::first();

所以它找不到类；

这是我的演示

文件1.php

<?php
namespace App;
class  Order{
public static function   first(){
echo "i am first";
  }
}

文件2.php

<?php
namespace App\Api\V1\Controllers;
include 'file1.php';

use App\Order ;
\App\Order::first();
// and you can do it by 
//  Order::first();  they are equal in this place 

当我运行命令php file2.php它回显我是第一个