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laravel - 编辑表单时未获取动态依赖列表值

问题描述

我正在使用动态相关下拉列表。子类别取决于类别。在我的编辑视图中，我想在我的表单中获取我之前选择的下拉值。但我只获得选定的类别..值。但不是子类别值...我不知道如何解决它

<div class="form-row col-md-12">
    <div class="form-group col-md-6" style="margin-left: 5px">
        <label for="inputError" class="control-label">Category</label>
        <select class="chosen-select form-control"   name="category">
            <option value="">Select category</option>
            @foreach ($categories as  $category)
                <option value="{{ $category->id }}" {{$product->category_id==$category->id?'selected':''}}>{{$category->category_name}}</option>
            @endforeach
        </select>
        {{$errors->first('category')}}
    </div>

    <div class="form-group col-md-6" style="margin-left: 5px" >
        <label for="inputError" class="control-label">Sub Category</label>
        <select class=" chosen form-control" id="subcategory" name="subcategory"></select>
        {{$errors->first('subcategory')}}
    </div>
</div>

这是我的ajax代码

<script type="text/javascript">
    $(document).ready(function() {
        $('select[name="category"]').on('change', function() {
            var stateID = $(this).val();
            console.log(stateID)
            if(stateID) {
                $.ajax({
                    url: '/myform/ajax/'+stateID,
                    type: "GET",
                    dataType: "json",
                    success:function(data) {
                        $('select[name="subcategory"]').empty();
                        $.each(data, function(key, value) {
                            $('select[name="subcategory"]').append('<option value="' + key + '">'+ value +'</option>');
                        });
                    }
                });
            }else{
                $('select[name="subcategory"]').empty();
            }
        });
    });
</script>

控制器

 public function myformAjax($id)

{

    $subcategories = DB::table("subcategories")

                ->where("category_id",$id)

                ->pluck("subcategory_name","id");

    return json_encode($subcategories);

}
public function edit($id)
{
     $categories = Category::latest()->get();
      $product = Product::findOrFail($id);

    return view('product.edit',compact('categories','product'));
}

路线

    Auth::routes();
Route::group(['middleware' => 'auth'], function (){
    Route::get('/home', 'HomeController@index')->name('home');
    Route::resource('group', 'GroupController');
    Route::resource('company', 'CompanyController');
    Route::resource('slider', 'SliderController');
    Route::resource('category', 'CategoryController');
    Route::resource('subcategory', 'SubcategoryController');
    Route::resource('product', 'ProductController');

});


Route::get('myform/ajax/{id}',array('as'=>'myform.ajax','uses'=>'HomeController@myformAjax'));

标签： laraveldropdown

解决方案

DB::table...将返回一个对象，但您需要有一个数组，因此您可以在调用之后使用 orall()或，如下所示：toArray()pluck(...)

$subcategories = DB::table("subcategories")->where("category_id",$id)->pluck("subcategory_name","id")->all();

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