Laravel 4 Jquery的AJAX級聯下拉

Question

我想基於在Laravel 4以前的選擇框值我想我有一個邏輯問題來填充選擇框：S

我的js`

$('#cat').change(function(){ 
     category_id = $(this).val(); 
     $('#secondcat').empty(); 
     $.ajax({ 
      type: 'GET', 
      url: '{{ URL::to('api/dropdown') }}', 
      data: 'category_id ='+category_id , 
      contentType: "application/json; charset=utf-8", 
      dataType: "json", 
      success:function(veri){ 
      $.each(veri,function(i,deger){ 

        $('#secondcat').append('<option value="'+deger.id+'">' +deger.name+ '<option>'); 

       }); // each 

      }, 
      error:function(x,hata){ 
       alert("Hata Oluştu" +hata); 
      } 

     }); // ajax 


    }); // change` 

路線

Route::get('api/dropdown','admin_BasvuruController@ajaxislemi'); 

我ajaxislemi（）方法在我BasvuruController：

public function ajaxislemi() { 

     $category_id = Input::get('category_id'); 

     return Kampanya::where('category_id','==',$category_id)->get(); 

    } 

在查看我的表格：

<select class="form-control" name="category_id" id="cat" > 
     <option value="" disabled selected> Please Select First </option> 
     @foreach($categories as $category) 
     <option value="{{ $category->id }}"> {{ $category->name }} </option> 
     @endforeach 
    </select> 



    <select name="kampanya_adi" class="form-control" id="secondcat" > 
     <option> </option> 
    </select> 

Answer 1

我相信這是你的問題：

return Kampanya::where('category_id','==',$category_id)->get(); 

你有雙 「==」。它應該是：

return Kampanya::where('category_id','=',$category_id)->get();