Laravel 4, - > withInput(); =未定义的偏移量:0

Question

我在这里和Laravel论坛上进行了长时间的搜索,但是我找不到这个问题的答案.->withInput()咳嗽起来Undefined offset: 0.

对于上下文:

调节器

public function getJobs()

        {
            $position_options = DB::table('jposition')->lists('friendly','id');
            $category_options = DB::table('jcategory')->lists('friendly','id');
            $location_options = DB::table('jlocation')->lists('friendly','id');



            $result = $query->get();
            return View::make('jobsearch.search', array('position_options' => $position_options, 'category_options' => $category_options, 'location_options' => $location_options))->withInput();

        }

视图

<form action="{{ action('JobsearchController@getJobs') }}" method="post">
  <div class="row">
    <div class="large-8 columns">
      <input type="text" name="realm" placeholder="Keywords/Skills" />
    </div>
    <div class="large-4 columns">
       {{ Form::select('category', $category_options , Input::old('category')) }}
    </div>
  </div>
  <div class="row">

    <div class="large-4 columns">
      {{ Form::select('location', $location_options , Input::old('location')) }}
    </div>


    <div class="large-4 columns">
      {{ Form::select('type', $position_options , Input::old('type')) }}
    </div>
    <div class="large-4 columns">
       <input type="submit" value="Search" style="width:100%; padding-top: .5rem;
padding-bottom: .5rem;" class="button border-btn" />
      </div>


</div>
</form>

现在根据文档应该没有问题,如果->withInput();删除页面加载正常.

最终目标是滚动我从上一个问题收到的答案来自db:raw的不期望的结果并且有一个页面加载"过滤"表单并在重新加载时显示相关结果并记住表单中的选择.

提前致谢.

更新:发表评论后我更新了控制器和路由,结果仍然相同:

routes.php文件

Route::get('jobs/search', 'JobsearchController@getSearch');

&

Route::post('jobs/search', 'JobsearchController@getJobs');

调节器

 public function getSearch()
        {
                    $position_options = DB::table('jposition')->lists('friendly','id');
            $category_options = DB::table('jcategory')->lists('friendly','id');
            $location_options = DB::table('jlocation')->lists('friendly','id');

            return View::make('jobsearch.search', array('position_options' => $position_options, 'category_options' => $category_options, 'location_options' => $location_options));
        }

        public function getJobs()

        {
            $position_options = DB::table('jposition')->lists('friendly','id');
            $category_options = DB::table('jcategory')->lists('friendly','id');
            $location_options = DB::table('jlocation')->lists('friendly','id');


            return View::make('jobsearch.search', array('position_options' => $position_options, 'category_options' => $category_options, 'location_options' => $location_options))->withInput();

        }

Answer 1

withInput()不像你想象的那样工作.它只是Redirect的功能,而不是View.

调用withInput($data)View具有完全不同的效果; 它将以下键值对传递给您的视图:( 'input' => $data您收到错误,因为您没有将任何数据传递给该函数)

要获得所需效果,请Input::flash()在进行观看之前致电,而不是致电withInput().这应该允许您使用Input::old()视图中的功能来访问数据.

或者,您可以简单地传递Input::all()到视图,并input[]在视图中使用该数组:

View::make(...)->withInput(Input::all());

这被翻译成

View::make(...)->with('input', Input::all());

至于你的评论,我建议这样做:

$position_options = DB::table('jposition')->lists('friendly','id');
$category_options = DB::table('jcategory')->lists('friendly','id');
$location_options = DB::table('jlocation')->lists('friendly','id');
$category = Input::get('category');
$location = Input::get('location');
$type = Input:: get('type'); 

$data = compact('position_options', 'category_options', 'location_options', 'category', 'type', 'location');

return View::make('jobsearch.search', $data);