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://code.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>
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);
}
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