Question

我想知道如何在表单中放置一个csrf令牌，以便删除它？

这是我的代码：

路线：

Route::delete('category_delete/{id}',['as'=>'category_delete','uses'=>'CategoryController@destroy']);

index.blade.php

@section('adminContent')
{!! Form::token() !!}
<div class="table-responsive art-content">
    <table class="table table-hover table-striped">
        <thead>
        <th> NAME</th>
        <th> ACTIONS</th>
        </thead>
        <tbody>
        @foreach($categoriesView as $category)
            <tr>
                <td>{!! $category->name!!}</td>
                <td>{!! link_to_route('categories.edit', '', array($category->id),array('class' => 'fa fa-pencil fa-fw')) !!}</td>
                <td><button type="button" id="delete-button" class="delete-button" data-url = "{!! url('category_delete')."/".$category->id !!}"><i class="fa fa-trash-o"></i></button>
            </td>
            </tr>
        @endforeach
        </tbody>
    </table>
    <div class="pagination"> {!! $categoriesView->render() !!}</div>
</div>

@stop

CategoryController：

 public function destroy($id,Category $category)
{

    $category = $category->find ( $id );
    $category->delete ();
    Session::flash ( 'message', 'The category was successfully deleted!.' );
    Session::flash ( 'flash_type', 'alert-success' );
}

如果我使用ajax，javascript或jquery，代码应该如何？

Answer 1

使用jquery我会做类似以下的事情。
在主视图的标题中添加以下行 <meta name="csrf-token" content="{{ csrf_token() }}" />
现在在你的javascript

function deleteMe(button)
   {
     // You might want to put the ajaxSetup function where it will always load globally.
     $.ajaxSetup({
         headers: {
             'X-CSRF-TOKEN': $('meta[name="csrf-token"]').attr('content')
         }
  });

   if(confirm("Are you sure?"))
   {
       $.get(
           'category_delete/'+button.id,
           function( response )
           {
               // callback is called when the delete is done.
               console.log('message = '+ response.message);
           }
       )
   }
  }

在index.blade.php中，让您的ID更具体 ... <td><button type="button" id=".$category->id" class="delete-button" data-url = "{!! url('category_delete')."/".$category->id !!}"><i class="fa fa-trash-o" onlick="deleteMe(this);></i></button> ...

在您的控制器中

public function destroy($id,Category $category){
     $category = $category->find ( $id );
     $category->delete ();

    return response()->json(
        'message' => 'Deleted', 
      )

}

Note：无需添加你认为Form::token 希望这会有所帮助.....

更新...........

如果您只使用laravel进行此操作，我建议您使用link而不是您正在使用的button。

在index.blade.php中，让您的ID更具体 ... <td><a href="category_delete/".$category->id class="delete-button" data-url = "{!! url('category_delete')!!}"><i class="fa fa-trash-o"></i></td> ...
在您的控制器中

public function destroy($id,Category $category){
    $category = $category->find ( $id );
    $category->delete ();

   return view('index');//important.

}

如果您希望链接看起来像按钮，请使用css或css框架，例如bootstrap 那应该是它。希望这会有所帮助。

laravel删除无效

1 个答案:

更新...........