我正在获取一个表单,允许用户更改他/她的密码,作为来自servlet的ajax响应。现在,我也在新表单中为按钮添加了ajax功能,但是当我单击更改密码按钮时,整个表单消失了,相反,我想再次单击更改时调用servlet密码即可。我确保从ajax调用接收响应的jsp文件 test.jsp 已经包含更改密码id的ajax逻辑 #changePswd
test.jsp的:
<%@ page language="java" contentType="text/html; charset=UTF-8"
pageEncoding="ISO-8859-1"%>
<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1">
<title>AJAX calls using Jquery in Servlet</title>
<script src="http://code.jquery.com/jquery-latest.js"></script>
<script>
$(document).ready(function() {
$('#submit1').click(function(event) {
event.preventDefault();
$.get('ActionServlet',{request:"form"},function(responseText) {
$('#welcometext').html(responseText);
});
});
$('#changePswd').click(function(event) {
event.preventDefault();
$.get('ActionServlet',{request:"action"},function(responseText) {
$('#content_progress').html(responseText);
});
});
});
$(document).ajaxStart(function(){
$('#content_progress').text("Loading...");
});
$(document).ajaxComplete(function(){
$('#content_progress').text("");
});
</script>
</head>
<body>
<form id="form1">
<h1>AJAX Demo using Jquery in JSP and Servlet</h1>
Enter your Name: <input type="text" id="user" /><br>
<a id="submit1" href="#">View Profile</a>
<a name="submit2" href="#">Course Details</a> <br />
<div id="content_progress"></div>
<div id="welcometext"></div>
</form>
</body>
</html>
的Servlet
package ajaxdemo;
import java.io.IOException;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
public class ActionServlet extends HttpServlet
{
private static final long serialVersionUID = 1L;
public ActionServlet() {
}
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException
{
String requestType=null;
String data = null;
requestType = request.getParameter("request").toString();
if(requestType.equals("form"))
{
data = "<form id = \"formChangePswd\"><table><tbody>"
+"<tr><td class=\"style1\">Old Password</td><td class=\"style2\"><input type=\"text\" id=\"oldPswd\" size=\"20\" class=\"textchange\" /></td></tr>"
+"<tr><td class=\"style1\">New Password</td><td class=\"style2\"><input type=\"text\" id=\"newPswd\" size=\"20\" class=\"textchange\"/></td></tr>"
+"<tr><td class=\"style1\">Confirm New Password</td><td class=\"style2\"><input type=\"text\" id=\"confirmPswd\" size=\"20\" class=\"textchange\"/></td></tr>"
+"<tr></tr><tr><td align=\"right\" class=\"style1\"><input type=\"reset\" id=\"reset\" value=\"Reset\" /></td><td class=\"style2\"><input type=\"submit\" id=\"changePswd\" value=\"Change Password\"/></td>"
+"</tr></tbody></table></form>";
}
else if(requestType.equals("action"))
{
data = "Your request is lodged, we will get back to you soon";
}
response.setContentType("text/html");
response.setCharacterEncoding("UTF-8");
response.getWriter().write(data);
}
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException
{
}
}
答案 0 :(得分:0)
如果我理解你的问题,你需要改变这一行:
$('#changePswd').click(function(event) {
到此:
$(document).on('click', '#changePswd', function(event) {
因为最初页面上不存在具有id changePswd的元素(并且稍后通过ajax / DOM操作添加),所以需要使用“on”方法而不是“click”方法。有关“on”方法的更多信息,请参阅jQuery API文档: