您可以在此处使用$ .ajax（）或$ .post。因为您已使用$ .ajax（）。请参考以下更正：
<!DOCTYPE html><html lang="en"><head> <title>SO question 4112686</title> <script src="http://pre.jquery.com/jquery-latest.min.js"></script> <script> $(document).ready(function() { $('#somebutton').click(function() { $.get('GetUserServlet', function(responseText) { $('#somediv').text(responseText); }); }); }); $("#somebutton").click(function(){ $.ajax({ url:'GetUserServlet', data:{name:'abc'}, type:'get', cache:false, success:function(data){ alert(data); $('#somediv').text(responseText); }, error:function(){ alert('error'); } } );}); </script></head><body> <button id="somebutton">press here</button> <div id="somediv"> </div></body>并且您的servlet应该是：
import java.io.IOException;import java.io.PrintWriter;import javax.servlet.*;import javax.servlet.http.*;public class GetUserServlet extends HttpServlet { public void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException { doPost(request, response); } public void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException { String text = "Update successfull"; //message you will recieve String name = request.getParameter("name"); PrintWriter out = response.getWriter(); out.println(name + " " + text); }
