问题描述
我的观点是 HTML 5.我正在使用 FormData 将 AJAX 2 POST 发送到 Servlet.在 servlet 内部，我试图读取请求参数.我看不到任何参数.但是，Google Chrome 开发控制台会显示请求负载.我怎样才能在 Servlet 代码中得到相同的结果?任何帮助将不胜感激.这是代码.
My view is HTML 5. I'm using FormData to make a AJAX 2 POST to a Servlet. Inside the servlet i'm trying to read request parameters. I can't see any parameters. However, Google Chrome Dev console shows the request payload. How can I get the same in Servlet code? Any help will be appreciated. Here's the code.
JS代码
var xhr = new XMLHttpRequest();
var formData = new FormData();
formData.append('firstName', 'ABC');
formData.append('lastName', 'XYZ');
xhr.open("POST", targetLocation, true);
xhr.send(formData);
Servlet 代码(两个参数都返回 null)
Servlet code (both parameters return null)
out.println("Hello! "+ request.getParameter("firstName")+ " "+ request.getParameter("lastName")+ ", thanks for sending your feedback." );
谷歌浏览器控制台
Content-Disposition: form-data; name="firstName"
XYZ
Content-Disposition: form-data; name="lastName"
ABC
推荐答案
HTML5 FormData API 发送 multipart/form-data 请求.它最初设计为能够通过 ajax 上传文件，新版本 2 XMLHttpRequest.以前的版本无法上传文件.
The HTML5 FormData API sends a multipart/form-data request. It's initially designed to be able to upload files by ajax, with the new version 2 XMLHttpRequest. Uploading files wasn't possible with the previous version.
request.getParameter() 默认只识别 application/x-www-form-urlencoded 请求.但是您正在发送 multipart/form-data 请求.您需要使用 @MultipartConfig 注释您的 servlet 类 以便你可以通过 request.getParameter() 获取它们.
The request.getParameter() by default recognizes application/x-www-form-urlencoded requests only. But you're sending a multipart/form-data request. You need to annotate your servlet class with @MultipartConfig so that you can get them by request.getParameter().
@WebServlet
@MultipartConfig
public class YourServlet extends HttpServlet {}
或者，如果您还没有使用 Servlet 3.0，请使用 Apache Commons FileUpload.有关这两种方法的更详细答案，请参阅:如何使用 JSP/Servlet 将文件上传到服务器?
Or, when you're still not on Servlet 3.0 yet, use Apache Commons FileUpload. For a more detailed answer on both approaches, see this: How to upload files to server using JSP/Servlet?
如果您根本不需要上传文件，请改用标准"XMLHttpRequest 方法.
If you don't need to upload files at all, use the "standard" XMLHttpRequest approach instead.
var xhr = new XMLHttpRequest();
var data = "firstName=" + encodeURIComponent(firstName)
+ "&lastName=" + encodeURIComponent(lastName);
xhr.open("POST", targetLocation, true);
xhr.setRequestHeader("Content-Type", "application/x-www-form-urlencoded");
xhr.send(data);
这样你的servlet就不再需要@MultipartConfig了.
This way you don't need @MultipartConfig on your servlet anymore.
- 如何使用 Servlet 和 Ajax?
- 通过 xmlHttpRequest 多部分发送文件
这篇关于HTML5 FormData 在 Java Servlet request.getParameter() 中返回 null的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持跟版网！