laravel upload files in many inputs

问题

I'm trying to upload files in 4 inputs files i get the solution from here but the problem the last file4 input file uploaded in all fields in database

in my blade form

{!! Form::file('file1', null,['class'=>'form-control']) !!}
{!! Form::file('file2', null,['class'=>'form-control']) !!}
{!! Form::file('file3', null,['class'=>'form-control']) !!}
{!! Form::file('file4', null,['class'=>'form-control']) !!}

in my controller

$input = $request->all();
    $files =[];
    if ($request->file('file1')) $files[] = $request->file('file1');
    if ($request->file('file2')) $files[] = $request->file('file2');
    if ($request->file('file3')) $files[] = $request->file('file3');
    if ($request->file('file4')) $files[] = $request->file('file4');
    foreach ($files as $file)
    {
        if(!empty($file)){
            $destinationPath = public_path() . '/uploads';
            $filename = $file->getClientOriginalName();
            $file->move($destinationPath, $filename);
        }

    }
    $model = new Project($input);
    $model -> file1 = $filename;
    $model -> file2 = $filename;
    $model -> file3 = $filename;
    $model -> file4 = $filename;
    $model->save();

回答1:

This is because you're accessing $filename outside of the foreach which will means only the last one is used.

You could do something like:

$input = $request->all();
$model = new Project($input);
$hasFiles = false;

foreach (range(1, 4) as $i) {
    $fileId = 'file' . $i;

    if ($request->hasFile($fileId)) {
        $hasFiles = true;

        $file = $request->file($fileId);

        $destinationPath = public_path() . '/uploads';
        $filename = $file->getClientOriginalName();
        $file->move($destinationPath, $filename);

        $model->$fileId = $filename;
    }
}

if ($hasFiles) {
    $model->save();
}

Hope this helps!

回答2:

If you want to go with this implementation then you should make the $filename an array because it will have the value of the last file (when you iterate).

if(!empty($file)){
     $destinationPath = public_path() . '/uploads';
     $filename[] = $file->getClientOriginalName();
     $file->move($destinationPath, $filename);
}

Then you can assign from that array:

$model = new Project($input);
$model -> file1 = isset($filename[0])?$filename[0]:null;
$model -> file2 = isset($filename[1])?$filename[0]:null;
$model -> file3 = isset($filename[2])?$filename[0]:null;
$model -> file4 = isset($filename[3])?$filename[0]:null;
$model->save();

Maybe a better checking can be made but the idea is to have an array with the uploaded file names.

Thoughts:

I would make a File model for the project and declare a hasMany relation in the Project and a belongsTo relation in the File model.

This would be the correct way to represent your data:

foreach($fileEntityList as $fileEntity){
    $project->files()->attach($fileEntity);
}

Check out "One To Many" in the docs

来源：https://stackoverflow.com/questions/45075572/laravel-upload-files-in-many-inputs