Question

当我将数据存储到数据库时，某处的某个字段会消失。

这是我的代码：

    public function artStore(FacilityRequest $request)
{
    $art = new Facility;

    $art->category()->attach($request->input('categories_list'));

    if(Input::hasfile('image'))
    {
    $request->file('image')->move(public_path('images/art/'), $request->file('image')->getClientOriginalName());

    $art->image = 'images/art/' . $request->file('image')->getClientOriginalName();
    }

    $art = Facility::create($request->all());

}

类别附加良好，但$art本身变为空。有人可以告诉我为什么吗？

这是错误：

SQLSTATE[23000]: Integrity constraint violation: 1048 Column 'facility_id' cannot be null (SQL: insert into `category_facility_facility` (`category_facility_id`, `facility_id`) values (2, ))

当我做dd（$ request）时会发生这种情况;

  +query: ParameterBag {#289 ▼
    #parameters: []
  }
  +server: ServerBag {#293 ▼
    #parameters: array:30 [▼
      "DOCUMENT_ROOT" => "C:\xampp\htdocs\uzdarbis\laravel\public"
      "REMOTE_ADDR" => "::1"
      "REMOTE_PORT" => "57855"
      "SERVER_SOFTWARE" => "PHP 5.6.12 Development Server"
      "SERVER_PROTOCOL" => "HTTP/1.1"
      "SERVER_NAME" => "localhost"
      "SERVER_PORT" => "8888"
      "REQUEST_URI" => "/create-art-work"
      "REQUEST_METHOD" => "POST"
      "SCRIPT_NAME" => "/index.php"
      "SCRIPT_FILENAME" => "C:\xampp\htdocs\uzdarbis\laravel\public\index.php"
      "PATH_INFO" => "/create-art-work"
      "PHP_SELF" => "/index.php/create-art-work"
      "HTTP_HOST" => "localhost:8888"
      "HTTP_CONNECTION" => "keep-alive"
      "HTTP_CONTENT_LENGTH" => "13360"
      "HTTP_CACHE_CONTROL" => "max-age=0"
      "HTTP_ACCEPT" => "text/html,application/xhtml+xml,application/xml;q=0.9,image/webp,*/*;q=0.8"
      "HTTP_ORIGIN" => "http://localhost:8888"
      "HTTP_UPGRADE_INSECURE_REQUESTS" => "1"
      "HTTP_USER_AGENT" => "Mozilla/5.0 (Windows NT 10.0; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/46.0.2490.86 Safari/537.36"
      "HTTP_CONTENT_TYPE" => "multipart/form-data; boundary=----WebKitFormBoundaryF4qxwnPWgFSBZl7r"
      "HTTP_REFERER" => "http://localhost:8888/facilities/add-new"
      "HTTP_ACCEPT_ENCODING" => "gzip, deflate"
      "HTTP_ACCEPT_LANGUAGE" => "lt,en-US;q=0.8,en;q=0.6,ru;q=0.4,pl;q=0.2"
      "HTTP_COOKIE" => "XSRF-TOKEN=eyJpdiI6IlByRUloN3djVVgxWDZ2MmpEN0lZRWc9PSIsInZhbHVlIjoiSU5lNkZieWcrY01KTG05Q3M5YmswcHNSQ3A0N05DQ2loTXVWbmszdnZldEl4TytMVkdZMTRYYUIwQjAxdEwwdmRjWFZTMnZtMzkzZUtMZHhnUE5nZWc9PSIsIm1hYyI6IjRjMDEzZTEzNzA0N2M1NWFhNmVlMTQ5Mjc5YmI0MjIzODc0NTRhZGJmZDkyNjJlMTk4ZTM0ZjEwMDc1YTRkMzEifQ%3D%3D; laravel_session=eyJpdiI6IlNPb3I3OHZqejVHOGg3V0FjWm9naGc9PSIsInZhbHVlIjoiSXkrcVhVRnJmXC9EVEVJOUJwR0hhYzZDVkZxcXh1V3NVUDM4RkdiNmQrZThwV0VPeklIZWxwSmNyeHp1NThBbm1rOXNLY0Q1S29xVDJNMmo4XC9RalBlQT09IiwibWFjIjoiMjNhMjE4NmUyMTUwZWE5YTJkYjhlMzU0ZWI3ODNhOTVmNzdiZTc1NWEwYjE5NzYzNDg0NzQ5N2U0Yjk3Njg1NCJ9"
      "REQUEST_TIME_FLOAT" => 1448714404.5915
      "REQUEST_TIME" => 1448714404
      "CONTENT_LENGTH" => "13360"
      "CONTENT_TYPE" => "multipart/form-data; boundary=----WebKitFormBoundaryF4qxwnPWgFSBZl7r"
    ]
  }

这是我的表格：

      <div class="formaIkelimo" style="width: 60%; margin: 0 auto"> 
    {!! Form::open(['url' => '/create-art-work/', 'files' => 'true']) !!}

    <div class="form-group">
            {!! Form::label('categories_list', 'Kategorija') !!}
            {!! Form::select('categories_list[]', $categories, null, ['id' => 'categories_list','class' => 'form-control']) !!}
    </div>

  <div class="form-group">

    <label>Photo</label>
    <input type="file" name="image" id="image" class="form-control">
  </div>    

    <div class="form-group">
    {!! Form::submit('Upload', ['class' => 'form-control btn btn-success']) !!}
    </div>

    {!! Form::close() !!}

    </div>

Answer 1

据我所知，你不是发送整数而是发送其他内容。我认为数据库需要一个整数。

你可以使用：

UICollectionView

查看请求包含的内容？

Answer 2

错误是因为您尝试将类别附加到任何内容。这导致错误：

$art = new Facility;
$art->category()->attach($request->input('categories_list'));

艺术是Facility的一个实例。它尚未进入数据库，因此它没有id。当您尝试附加类别时，facility_id为空。这导致了你的错误。处理此问题的最佳方法是在创建设施后将其附加到最后：

public function artStore(FacilityRequest $request)
{
    $art = new Facility;

    if(Input::hasfile('image'))
    {
         $request->file('image')->move(public_path('images/art/'), $request->file('image')->getClientOriginalName());

         $art->image = 'images/art/' . $request->file('image')->getClientOriginalName();
    }

    $art = Facility::create($request->all());

    $art->category()->attach($request->input('categories_list'));
}

这是假设您的请求输入正确。下次，您应该dd($request->all())而不是$request。

修改：我还意识到您的image属性无法使用您当前的方法正确保存。我建议这样做，这也有点清洁：

public function artStore(FacilityRequest $request)
{
    if(Input::hasFile('image'))
    {
         $request->file('image')->move(public_path('images/art/'), $request->file('image')->getClientOriginalName());

         $request->merge(['image' => 'images/art/' . $request->file('image')->getClientOriginalName()]);
    }

    $art = Facility::create($request->all());

    $art->category()->attach($request->input('categories_list'));
}

将文件移动到指定区域后，merge方法基本上会用新值覆盖image属性。这样，您可以在最后一行创建Facility对象。然后，您附加类别。

在Laravel 5中存储数据时获取null

2 个答案: