laravel 5.2 |上傳文件 - 對空

Question

調用一個成員函數getClientOriginalName（）我試圖上傳資料圖片PICT，但我得到的錯誤

「上的空調用一個成員函數getClientOriginalName（）」這是我的方法：

$data = $request->input('fotodosen'); 
    $photo = $request->file('fotodosen')->getClientOriginalName(); 
    $destination = base_path() . '/public/uploads'; 
    $request->file('fotodosen')->move($destination, $photo); 
    $data['fotodosen'] = $photo; 
    Dosen::create($data); 

創建：

{!! Form::open(array('fotodosen'=>'create', 'method'=>'POST', 'files'=>true, 'url'=>'uploads')) !!} 
      {!! Form::file('image') !!} 
        <div class="form-group"> 
         <div class="col-md-6 col-md-offset-4"> 
          <button type="submit" class="btn btn-primary"> 
           <i class="fa fa-btn fa-user"></i> Register 
          </button> 
          {!! Form::close() !!} 

已經編輯方法：

$photo = $request->file('fotodosen')->getClientOriginalName($photo); 

仍然有這個錯誤。我錯過了什麼？

UPDATE：

public function store(CreateDosenRequest $request) 

{ 



    $user = User::create([ 
     'name' => $request->input('name'), 
     'username' => $request->input('username'), 
     'email' => $request->input('email'), 
     'password' => $request->input('password'), 
     'admin' => $request->input('admin'), 
    ]); 

     $dosen = Dosen::create([ 
     'iddosen' => $request->input('iddosen'), 
     'nipy' => $request->input('nipy'), 
     'namadosen' => $user->name, 
     'user_id' => $user->id, 
     'alamatdosen' => $request->input('alamatdosen'), 
     'notelpdosen' => $request->input('notelpdosen'), 
     'tempatlahirdosen' => $request->input('tempatlahirdosen'), 
     'tanggallahirdosen' => $request->input('tanggallahirdosen'), 
     'agamadosen' => $request->input('agamadosen'), 

    ]); 
     if ($request->hasFile('image')) { 
      $data = $request->input('image'); 
      $photo = $request->file('image')->getClientOriginalName(); 
      $destination = public_path() . '/uploads/'; 
      $request->file('image')->move($destination, $photo); 
      $data['fotodosen'] = $photo; 
      Dosen::create($data); 
} 

Answer 1

你有文件名image嘗試使用image代替fotodosen

$photo = $request->file('image')->getClientOriginalName(); 

的完整代碼

$data = $request->input('image'); 
$photo = $request->file('image')->getClientOriginalName(); 
$destination = base_path() . '/public/uploads'; 
$request->file('image')->move($destination, $photo); 

您可以檢查在t他文件一樣，

if ($request->hasFile('image')) { 
    // your code here 
} 

從Http Requests和文章file upload in laravel 5

Answer 2

我有同樣的問題，所以我使用的HTML表單屬性enctype="multipart/form-data"

解決它實例

<form name="le_form" action="/ft" method="POST" enctype="multipart/form-data"> 

欲瞭解更多信息見

• http://php.net/manual/en/features.file-upload.post-method.php

• https://developer.mozilla.org/en-US/docs/Web/API/HTMLFormElement/enctype