使用多部分发送附加数据 [英] Sending additional data with multipart

问题描述

我正在使用 apache-commons-fileupload 从 client 获取文件到 server.(使用 JSP 和 Servlet).

I am using apache-commons-fileupload to get file from client to the server.(using JSP and Servlet).

JSP/HTML

<form method="POST" action="GetFile" enctype="multipart/form-data">
<input type="file" name="datafile">
<input type="text" name="text1">
<input type="submit" value="Next">
</form>

Servlet:获取文件

System.out.println(request.getParameter("text1"));

我能够将文件上传到服务器，但我无法在 servlet 中获取 text1 的值(我得到 nullservlet 中 text1 的 值)，我需要表单中的这个 textfield 来提交一些额外的信息，同时将它上传到服务器.

I am able to upload the file to the server, but I am not able to get the value of text1 in the servlet (I am getting null value of text1 in the servlet), I need this textfield in the form to submit some additional information while uploading it to the server.

• Is enctype="multipart/form-data" 表单选项不允许其他表单数据要提交?如果它不允许，那么我必须将这个额外的 textfield 发送到 server 的其他选项是什么.
• 或者我的代码还有其他问题吗?

• Is enctype="multipart/form-data" option of form doesn't allow other form data to be submited? if it doesn't allow it then what are the other options I have to send this additional textfield to the server.
• Or is there any other problem in my code?

推荐答案

Is enctype="multipart/form-data" 表单选项不允许其他表单数据要提交?如果它不允许，那么是什么其他选项我必须将此附加文本字段发送到服务器.

Is enctype="multipart/form-data" option of form doesn't allow other form data to be submited? if it doesn't allow it then what are the other options I have to send this additional textfield to the server.

不，使用 enctype="multipart/form-data" 没有问题.您可以获取其他字段，然后以这种形式提交.

No there is no issue with using enctype="multipart/form-data". You can get other fields then file in such form.

或者我的代码还有其他问题吗?

Or is there any other problem in my code?

是的，就目前而言.在使用 enctype="multipart/form-data" 时，您不能通过使用 request.getParameter(name); 直接获取参数.使用它时，表单字段不能作为请求的参数使用，它们包含在流中，因此您无法以正常方式获取它.您可以在 使用 commons-fileupload 的文档中找到一种方法#处理上传的项目.

Yes, as for now. While using enctype="multipart/form-data" you can not directly get parameters by using request.getParameter(name);. While using it, form fields aren't available as parameter of the request, they are included in the stream, so you can not get it the normal way. You can find a way to do this in the docs of using commons-fileupload#Processing the uploaded items.

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