php - Laravel 5 返回 JSON 或 View 取决于是否使用 ajax

Question

我想知道是否有一种神奇的方法可以使用这种情况:

如果我通过 AJAX 请求调用页面， Controller 返回一个 JSON 对象，否则它返回一个 View ，我试图在我的所有 Controller 上执行此操作，而无需更改每个方法。

例如我知道我可以做到:

if (Request::ajax()) return compact($object1, $object2);
else return view('template', compact($object, $object2));

但我有很多 Controller /方法，我更愿意更改基本行为，而不是花时间更改所有这些。有什么想法吗？

Answer 1

最简单的方法是创建一个在所有 Controller 之间共享的方法。

示例:

这是所有其他 Controller 扩展的 Controller 类:

<?php namespace App\Http\Controllers;

use Illuminate\Routing\Controller as BaseController;

abstract class Controller extends BaseController
{
    protected function makeResponse($template, $objects = [])
    {
        if (\Request::ajax()) {
            return json_encode($objects);
        }

        return view($template, $objects);
    }
}

这是扩展它的 Controller 之一:

<?php namespace App\Http\Controllers;

class MyController extends Controller
{
    public function index()
    {
        $object = new Object1;
        $object2 = new Object2;

        return $this->makeResponse($template, compact($object, $object2));
    }
}

Laravel 5+ 更新

<?php

namespace App\Http\Controllers;

use Illuminate\Foundation\Bus\DispatchesJobs;
use Illuminate\Routing\Controller as BaseController;
use Illuminate\Foundation\Validation\ValidatesRequests;
use Illuminate\Foundation\Auth\Access\AuthorizesRequests;

class Controller extends BaseController
{
    use AuthorizesRequests, DispatchesJobs, ValidatesRequests;

    protected function makeResponse($request, $template, $data = [])
    {
        if ($request->ajax()) {
            return response()->json($data);
        }

        return view($template, $data);
    }
}

<?php

namespace App\Http\Controllers;

use Illuminate\Http\Request;

class MyController extends Controller
{
    public function index(Request $request)
    {
        $object = new Object1;
        $object2 = new Object2;

        return $this->makeResponse($request, $template, compact($object, $object2));
    }
}