如何修复“未定义的变量：gambar（查看：C：\ xampp \ htdocs \ apmt \ resources \ views \ upload.blade.php）

Question

我不明白为什么gambar上foreach的变量未定义。我也是PHP Laravel 5.8的新手。

（upload.blade.php）

<table class="table table-bordered table-striped">
    <thead>
        <tr>
            <th width="1%">File</th>
            <th>Keterangan</th>
            <th width="1%">OPSI</th>
        </tr>
    </thead>
    <tbody>

        @foreach ($gambar as $g)`here is the undefined variable`
          <tr>
              <td><img width="150px" src="{{ url('/data_file/'.$g->file) }}"></td>
              <td>{{$g->keterangan}}</td>
              <td><a class="btn btn-danger" href="/upload/hapus/{{ $g->id }}">HAPUS</a></td>
          </tr>
        @endforeach
    </tbody>
</table>

UploadController.php

    public function upload(){
        $gambar = Gambar::get();
        return view('upload',['gambar' => $gambar]);
    }   

    public function proses_upload(Request $request){
        $this->validate($request, [
            'file' => 'required|file|image|mimes:jpeg,png,jpg|max:2048',
            'keterangan' => 'required',
        ]);

        // menyimpan data file yang diupload ke variabel $file
        $file = $request->file('file');

        $nama_file = time()."_".$file->getClientOriginalName();

                // isi dengan nama folder tempat kemana file diupload
        $tujuan_upload = 'data_file';
        $file->move($tujuan_upload,$nama_file);

        Gambar::create([
            'file' => $nama_file,
            'keterangan' => $request->keterangan,
        ]);

        return redirect()->back();
    }

我希望它会显示我文件（data_file）中的图像。我上传的所有图片都工作正常。只是无法显示在表格中。

Answer 1

您错过了在gambar函数中传递proses_upload的情况，我想您正在将图像插入upload.blade.php

public function proses_upload(Request $request){
$this->validate($request, [
    'file' => 'required|file|image|mimes:jpeg,png,jpg|max:2048',
    'keterangan' => 'required',
]);

// menyimpan data file yang diupload ke variabel $file
$file = $request->file('file');

$nama_file = time()."_".$file->getClientOriginalName();

        // isi dengan nama folder tempat kemana file diupload
$tujuan_upload = 'data_file';
$file->move($tujuan_upload,$nama_file);

Gambar::create([
    'file' => $nama_file,
    'keterangan' => $request->keterangan,
]);
$gamber = Gambar::get();
return view('upload',['gambar' => $gambar]);

}

Answer 2

请编辑您的帖子，并在问题中添加控制器代码。 但我认为。 $ gambar未在您的控制器中正确传递。 该步骤可能会有所帮助：

1. 签入您的控制器，是否已将$ gambar正确传递到您的视图。
laravel中的

2. 。您可以使用compact或with()方法来传递变量。 控制器中的示例：

   $gambar = 'path/to/file/gambar.png'
   return view('index',compact('gambar')) // it will passed $gambar in index.blade
   

或

$gambar = 'path/to/file/gambar.png'
return view('index')->with('gambar',$gambar); // it's also passed $gambar in index.blade

Answer 3

public function upload()
{
$gambar = Gambar::get();
return view('upload',compact('gambar'));
OR
return viw('upload')->with('gambar',$gambar);
}   

public function proses_upload(Request $request){
$this->validate($request, [
    'file' => 'required|file|image|mimes:jpeg,png,jpg|max:2048',
    'keterangan' => 'required',
]);

// menyimpan data file yang diupload ke variabel $file
$file = $request->file('file');

$nama_file = time()."_".$file->getClientOriginalName();

        // isi dengan nama folder tempat kemana file diupload
$tujuan_upload = 'data_file';
$file->move($tujuan_upload,$nama_file);

Gambar::create([
    'file' => $nama_file,
    'keterangan' => $request->keterangan,
]);

return redirect()->route('upload-view');
} 

在web.php

  Route::get('upload',UploadController@upload)->name('upload-view');

您认为

  <table class="table table-bordered table-striped">
            <thead>
                <tr>
                    <th width="1%">File</th>
                    <th>Keterangan</th>
                    <th width="1%">OPSI</th>
                </tr>
            </thead>
            <tbody>

                @foreach($gambar as $g)
                <tr>
                    <td><img width="150px" src="{{ url('/data_file/'.$g->file) }}"></td>
                    <td>{{$g->keterangan}}</td>
                    <td><a class="btn btn-danger" href="/upload/hapus/{{ $g->id }}">HAPUS</a></td>
                </tr>
            @endforeach
            </tbody>
        </table>

Answer 4

请执行此操作

$gambar = Gambar::all();
return view('upload',compact('gambar'));

首先在控制器中返回该变量，以检查其返回的值，就像我在下面所做的那样，如果返回，则使用：

$gambar = Gambar::all();
return $gambar;