如何使用 Laravel 5.1 在表中保存上传的文件名
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/33150998/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
How to save uploaded file name in table using Laravel 5.1
提问by dollar
I need help in saving uploaded file name in database table using laravel 5.1.
我需要帮助使用 laravel 5.1 在数据库表中保存上传的文件名。
My Controller code for saving Image details
用于保存图像详细信息的我的控制器代码
public function store(Request $request)
{
if($request->hasFile('img_filename'))
{
$destinationPath="offerimages";
$file = $request->file('img_filename');
$filename=$file->getClientOriginalName();
$request->file('img_filename')->move($destinationPath,$filename);
}
$input=$request->all();
Offer_image::create($input);
return redirect('offerimage');
}
My view code for accepting image
我的接受图像的查看代码
{!! Form::open(array('route'=>'offerimage.store','role'=>'form','files'=>true)) !!}
<div class="box-body">
<div class="form-group">
{!! Form::label('img_name','Name') !!}
{!! Form::text('img_name', $value = null, $attributes = array('class'=>'form-control','id'=>'img_name','required')) !!}
</div>
<div class="form-group">
{!! Form::label('img_description','Description') !!}
{!! Form::textarea('img_description', $value = null, $attributes = array('class'=>'form-control','id'=>'img_description','required')) !!}
</div>
<div class="form-group">
{!! Form::label('img_filename','Upload Image') !!}
{!! Form::file('img_filename') !!}
</div>
{!! Form::hidden('status',$value='active') !!}
</div><!-- /.box-body -->
<div class="box-footer">
{!! Form::submit('Submit',$attributes=array('class'=>'btn btn-primary')) !!}
</div>
{!! Form::close() !!}
This controller code to store image working properly, but where i am trying to save image file name to table , this code is storing filepath to database table.
这个用于存储图像的控制器代码工作正常，但是在我尝试将图像文件名保存到表的地方，此代码将文件路径存储到数据库表。
As I am using direct create() method to store the request object in table, I don't know how do I store file name instead of path.
由于我使用直接 create() 方法将请求对象存储在表中，因此我不知道如何存储文件名而不是路径。
回答by darthsoup
The Problem is that your Request Data hasn't changed while you uploaded the picture. So img_filenamestill contains tmpdata.
问题是您上传图片时您的请求数据没有改变。所以img_filename仍然包含tmpdata。
You can try this:
你可以试试这个：
$input = $request->all();
$input['img_filename'] = $filename;
回答by fico7489
Code that works for me :
对我有用的代码：
$updir = 'images/';
$img_name = 'image.jpeg';
Request::file('img_filename')->move($updir, $img_name);
