Question

我有两个方法show和index我在索引方法上有一个项目列表，当用户点击它时，会将她带到包含属于该id的一些数据的另一个页面。我没有这样做，而是想使用jquery在同一页面上加载数据。我在index.blade.php视图中有以下内容。我如何在laravel中实现这一目标？

  @foreach ($categories as $category)

    <div class="body">
    <h4><a style="text-decoration: none; " href="{{ URL::route('category.show', $category->id) }}">{{$category->name}}</a></h4>

    </div>
   @endforeach

<?php
 namespace App\Http\Controllers;
 use Illuminate\Http\Request;
 use illuminate\HttpResponse;
 use App\Http\Requests\todolistRequest;
 use App\Http\Requests\CreateCategoryRequest;
 use App\Http\Requests;
 use App\Companylist;
 use App\Category;
 use  Illuminate\Support\Facades\DB;
 class CategoryController extends Controller
 {

public function create(){
return view('category.create');
 }

 public function index(){
 $categories=Category::all();
 return view('category.index',compact('categories'));
 }
  public function store(CreateCategoryRequest $request){
 $category = new Category($request->all());
 $category->save();

  return \Redirect::route('category.create')->with('message',      'Your list has been created!'); 
  }
  public function show($id)
  {
  $category = Category::findOrFail($id)->companylist()->get();
  $cat=Category::findOrFail($id);
// this my route
   Route::resource('category','CategoryController');

      return                           view('category.show')->with('category',                          $category)->with('cat',$cat);
 }

// }

Answer 1

使用引导模式或弹出窗口。根据您的需求设计category.show视图。调用ajax并将返回的视图响应附加到该模态体并显示它，以便它与您想要的视图完全相同。

示例代码用于替换，而ajax用于

这是替换模态

<!-- Show Modal content-->
<div class="modal-content">
  <div class="modal-header">
    <button type="button" class="close" data-dismiss="modal">&times;</button>
    <h4 class="modal-title">Show data</h4>
  </div>
  <div class="modal-body replace">

  </div>
</div>

这将是你的ajax方法

      $('.showModalButton').on('click',function(e){
        var id = $(this).data('id'); // get id of resource you want to show
        $('.replace').empty(); // 
        $.ajax({
            url: 'your route/'+id,
            type: "GET",

            success:function(data) {
                $('.replace').html(data);
                },
            error:function(jQXHR, textStatus, errorThrown) {
                console.log(jQXHR);
                console.log(textStatus);
                console.log(errorThrown);
                }
        }); //end Ajax call

    });

这将是您在控制器中的show方法

public function show($id)
   {
     $category = Category::findOrFail($id)->companylist()->get();
     $cat=Category::findOrFail($id);
     // this my route
     Route::resource('category','CategoryController');

      return view('category.show')->with('category',$category)->with('cat',$cat);
    }

设计category.show但是你要显示你的数据。

如何使用jquery在索引方法上加载laravel中的show方法的内容？

1 个答案: