laravel在许多输入中上传文件 [英] laravel upload files in many inputs
本文介绍了laravel在许多输入中上传文件的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我正在尝试上传4个输入文件中的文件,我从中获得了解决方案 此处但是问题出在数据库的所有字段中最后上传的file4输入文件
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
以我的刀片形式
{!! 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']) !!}
在我的控制器中
$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;
$model->save();
推荐答案
这是因为您正在foreach之外访问$filename
,这意味着仅使用了最后一个.
This is because you're accessing $filename
outside of the foreach which will means only the last one is used.
您可以执行以下操作:
$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();
}
希望这会有所帮助!
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