laravel - 尝试使用laravel获取非对象的属性 'catid'

Question

这是我的代码

public function edit($id) {
    $categories = Category::all();
    $category = Category::find($id);
    return view('categories.edit', compact('categories', 'category'));

}

以下是我的编辑表单

{!! Form::open(['url' => ['categories', $category->catid], 'method' => 'put']) !!}
    <div class="form-group padding-t-20 {{ $errors->has('name') ? 'has-error' : '' }}">
        {{ Form::label('name', 'Name') }}
        {{ Form::text('name', $category->name, ['class' => 'form-control', 'placeholder' => 'Category']) }}
        <span class="text-danger">{{ $errors->has('name') ? $errors->first('name') : '' }}</span>
    </div>
    <div class="text-center padding-t-20">
            {{ Form::submit('Update', ['class' => 'btn btn-info btn-fill btn-wd']) }}
    </div>
{!! Form::close() !!}

它工作正常，但是当我使用如下所示的数据库外观时

 public function edit($id) {
        $categories = DB::select('select * from tbl_menu_category');
        $category = DB::select('select * from tbl_menu_category where catid = ?', [$id]);
        return view('categories.edit', compact('categories', 'category'));
}

尝试获取非对象的属性“catid”时出现错误

Answer 1

当您使用 DB::select() 查询数据库时，即使您试图从数据库中仅获取记录，它也会以数组形式返回结果。

所以解决方案是:

public function edit($id) {
  $categories = DB::select('select * from tbl_menu_category');
  $category = DB::select('select * from tbl_menu_category where catid = ?', [$id])[0]; <--- grab the first index of the array returned.
  return view('categories.edit', compact('categories', 'category'));
}

希望对你有帮助:)