在Laravel中

Question

大家好我的项目需要帮助laravel如何在后端处理这个问题？

如果你有多个输入怎么样：

Photo 1: 
<input type="file" name="name">
Photo 2: 
<input type="file" name="name">

    $file_upload = [];
    $("input[name=name]").each(function()){
      $upload_val = $(this).prop('files')[0];
      $file_upload.push($upload_val);
    }


   var data_post = new FormData();	
   data_post.append('uploads',$file_upload);
   $.ajax({
		url: "app/upload",
		type: "POST",
		processData: false,
		contentType: false,
		data: data_post,
		success:function(res){
          console.log(res)
        }
   });

Answer 1

我相信你可以做点什么

您的HTML

<input type="file" value="photos[]">
<input type="file" value="photos[]">

你的后端

$photos = $request::file('photos')
foreach($photos as $photo) {
    // your code
}