jsp - 使用 ServletFileUpload 的 parseRequest 上传文件？

Question

这个问题在这里已经有了答案:





How to upload files to server using JSP/Servlet?

(14 个回答)


5年前关闭。




我上传了我浏览的文件 input type="file"在我的网络应用程序中。问题是我得到了 FileItem列表大小为 0 虽然我可以在下面看到所有上传的文件信息
request -> JakartaMutltiPartRequest -> 文件属性

这是读取文件的java代码

public InputStream parseRequestStreamWithApache(HttpServletRequest request)
  throws FileUploadException, IOException {
  InputStream is = null;
  FileItemFactory factory = new DiskFileItemFactory();
  ServletFileUpload upload = new ServletFileUpload(factory);
  List items = upload.parseRequest(request);
  // here the item size is 0 ,i am not sure why i am not getting my file upload in browser with type="file"
  // but If inspect request in debugger i can see my file realted info in request--->JakartaMutltiPartRequest----->files attribute
  Iterator iter = items.iterator();
  while (iter.hasNext()) {
    FileItem item = (FileItem) iter.next();

    if (!item.isFormField()) {
      is = item.getInputStream();
    }
  }

  return is;
}

编辑:

这是 JSP 部分:

<form NAME="form1" action="customer/customerManager!parseRequestStreamWithApache.action" ENCTYPE="multipart/form-data"   method="post" >
     <TABLE >
         <tr>
              <th>Upload File</th>
                  <td>
                   <input name="fileUploadAttr" id="filePath"  type="file" value="">
                 </td>
                  <td > 
                 <Input type="submit" value ="uploadFile"/>
                  </td>
          </tr>
    </TABLE>
</form>

Answer 1

正如我在对同一问题的评论中所说的，您之前发布过，这很可能是因为您之前已经解析了请求。
这些文件是请求正文的一部分，您只能解析一次。

更新:

我通常以这种方式使用 commons-upload:

if (ServletFileUpload.isMultipartContent(request)) {
    ServletFileUpload fileUpload = new ServletFileUpload();
    FileItemIterator items = fileUpload.getItemIterator(request);
    // iterate items
    while (items.hasNext()) {
        FileItemStream item = items.next();
        if (!item.isFormField()) {
            is = item.openStream();
        }
    }
}