我有一个表单,用户应至少选择一个要上传的文件。我有三个文件输入字段,如下所示:
<div class="form-group col-lg-4">
{!! Form::label('file1', 'Select file 1', ['class' => 'control-label']) !!}
{!! Form::file('files[]', ['id'=>'file1']) !!}
</div>
<div class="form-group col-lg-4">
{!! Form::label('file2', 'Select file 2', ['class' => 'control-label']) !!}
{!! Form::file('files[]', ['id'=>'file2']) !!}
</div>
<div class="form-group col-lg-4">
{!! Form::label('file3', 'Select file 3', ['class' => 'control-label']) !!}
{!! Form::file('files[]', ['id'=>'file3']) !!}
</div>
我应该验证表单请求中是否存在至少一个文件和 MIME 类型。然后在相关表单 Controller 的store方法中,将原始文件名存储在对应的三个数据库字段(即file1、file2、file3)中。
我该如何实现这个?
最佳答案
经过一番查找,终于找到了解决办法。首先,我将 View 修改为如下所示:
<div class="form-group col-lg-4">
{!! Form::label('file1', 'Select file 1', ['class' => 'control-label']) !!}
{!! Form::file('file1', ['id'=>'file1']) !!}
</div>
<div class="form-group col-lg-4">
{!! Form::label('file2', 'Select file 2', ['class' => 'control-label']) !!}
{!! Form::file('file2', ['id'=>'file2']) !!}
</div>
<div class="form-group col-lg-4">
{!! Form::label('file3', 'Select file 3', ['class' => 'control-label']) !!}
{!! Form::file('file3', ['id'=>'file3']) !!}
</div>
然后在 Controller 中我使用了您建议的代码:
$files =[];
if ($request->file('file1')) $files[] = $request->file('file1');
if ($request->file('file2')) $files[] = $request->file('file2');
if ($request->file('file3')) $files[] = $request->file('file3');
foreach ($files as $file)
{
if(!empty($file)){
$filename=$file->getClientOriginalName();
$file->move(
base_path().'/public/uploads/', $filename
);
}
}
关于php - 拉维尔 5.1 : How to upload multiple files from three different file input fields?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/31936336/