Laravel 4 - ErrorException未定義變量

Question

嘗試將模型綁定到表單以調用更新函數，但未找到模型。

{{ Form::model($upload, array('url' => array('uploads/update', $upload->id), 'files' => true, 'method' => 'PATCH')) }} 

控制器來獲取編輯觀點

public function getEdit($id) 
{ 
$upload = $this->upload->find($id); 

if (is_null($upload)) 
{ 
    return Redirect::to('uploads/alluploads'); 
} 

$this->layout->content = View::make('uploads.edit', compact('uploads')); 
} 

控制器做更新

public function patchUpdate($id) 
{ 
    $input = array_except(Input::all(), '_method'); 

    $v = Validator::make($input, Upload::$rules); 

    if ($v->passes()) 
    { 
    $upload = $this->upload->find($id); 
    $upload->update($input); 

    return Redirect::to('uploads/show', $id); 
    } 

    return Redirect::to('uploads/edit', $id) 
    ->withInput() 
    ->withErrors($v) 
} 

錯誤，我得到

ErrorException 
Undefined variable: upload (View: /www/authtest/app/views/uploads/edit.blade.php) 

Answer 1

如果你沒有約束力的模型然後從控制器傳遞它當顯示/加載要編輯的形式，例如：

public function getEdit($id) 
{ 
    $upload = $this->upload->find($id); 
    if (is_null($upload)) 
    { 
     return Redirect::to('uploads/alluploads'); 
    } 
    $this->layout->content = View::make('uploads.edit', compact('upload')); //<-- 
} 

更新：您應該使用compact('upload')不uploads，因爲變量是$upload。