我开始学习与jQuery的Ajax,我试了很多谷歌搜索,我不能得到这个测试代码工作,你能告诉我什么是我做错了?会发生什么是ajax不会创建表。这里是我的代码:我不能得到这个简单的Ajax代码工作
的JSP:
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<script type="text/javascript" src="js/jquery.js"></script>
<title>My First Web App</title>
<script type="text/javascript">
$(document).ready(function{
$.ajax({
type: "GET",
url: "users",
dataType: "xml",
success: function(xml){
$("#content").append("<table>");
$(xml).find("user").each(function(){
var firstName = $(this).find("firstName").text();
var lastName = $(this).find("lastName").text();
var password = $(this).find("password").text();
var email = $(this).find("email").text();
$("#content").append("<tr>");
$("#content").append("<td>" + firstName + "</td>");
$("#content").append("<td>" + lastName + "</td>");
$("#content").append("<td>" + password + "</td>");
$("#content").append("<td>" + email + "</td>");
$("#content").append("</tr>");
});
$("#content").append("</table>");
}
});
});
</script>
</head>
<body>
<div id="content"></div>
</body>
该servlet:
@WebServlet("/users")
public class users extends HttpServlet {
private static final long serialVersionUID = 1L;
/**
* @see HttpServlet#HttpServlet()
*/
public users() {
super();
// TODO Auto-generated constructor stub
}
/**
* @see HttpServlet#doGet(HttpServletRequest request, HttpServletResponse response)
*/
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
response.setContentType("text/xml;charset=UTF-8");
usuarioDAO uDAO = new usuarioDAO();
response.getWriter().write(uDAO.getAllUsers());
}
usuarioDAO:
public String getAllUsers()
{
String xml = "";
xml += "<?xml version=\"1.0\" encoding=\"UTF-8\"?>";
xml += "<users>";
try
{
getAllUsers = con.prepareStatement("SELECT * FROM users");
synchronized(getAllUsers)
{
ResultSet res = getAllUsers.executeQuery();
while (res.next())
{
xml += "<user>";
xml += "<firstName>" + res.getString("firstName") + "</firstName>";
xml += "<lastName>" + res.getString("lastName") + "</lastName>";
xml += "<password>" + res.getString("password") + "</password>";
xml += "<email>" + res.getString("email") + "</email>";
xml += "</user>";
}
}
getAllUsers.close();
}
catch (Exception ex)
{
System.out.println(ex);
}
xml += "</users>";
return xml;
}
而且仅此而已,你能不能请告诉我我做了什么错了吗?
定义 “不工作”。它以什么方式失败?是否有某种错误?例外?代码中的哪一点与预期行为有偏差?在这一点上,预期与观察行为是什么?那时相关对象的状态是什么? – David
你确定你的ajax调用的“url”是正确的吗? – MilkyWayJoe
也许你忘了在这里调用response.close()或response.flush()?如果它不会帮助,请您发布一些额外的细节,可能是服务器日志或萤火虫日志? –