使用ServletFileUpload的parseRequest上传文件？

  • Z时代
  • 2024-01-10
  • 分类：问答

我上传了inputtype="file"在Web应用程序中浏览的文件。问题是我可以将FileItem列表大小设为0，尽管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;
}

<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>

回答：

正如我在您之前发布的对同一个问题的评论中所说，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。

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();
        }
    }
}

以上是 使用ServletFileUpload的parseRequest上传文件？ 的全部内容， 来源链接： utcz.com/qa/421368.html

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