简单的ajax登陆验证

来源：互联网发布：郑州掌沃软件科技编辑：程序博客网时间：2024/04/29 20:28

1.login.jsp

<%@ page language="java" contentType="text/html; charset=utf-8"
pageEncoding="utf-8"%>
<!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>login</title>
<script type="text/javascript">
function createAjax() {
var xmlHttp;
try {
xmlHttp = new XMLHttpRequest();
} catch (e) {
try {
xmlHttp = new ActiveXObject("Msxm12.XMLHTTP");
} catch (e) {
xmlHttp = new ActiveXObject("Microsoft.XMLHTTP");
}
}
return xmlHttp;
};
function checkLogin() {
var loginName = document.getElementById("username").value;
var xmlHttp = createAjax();
xmlHttp.onreadystatechange = function() {
if (xmlHttp.readyState == 4) {
if (xmlHttp.status == 200) {
var data = xmlHttp.responseText;
alert(document.getElementById("suggest"));
document.getElementById("suggest").innerHTML = data;
}
}
};
xmlHttp.open("post", "checkLogin", true);
xmlHttp.setRequestHeader("Content-Type",
"application/x-www-form-urlencoded");
xmlHttp.send("loginName=" + loginName);
};
</script>
</head>
<body>
<form action="">
用户名：<input type="text" name="username" id="username"
onblur="checkLogin()" /><span id="suggest"></span><br>
密   码：<input type="password" name="password"
id="password" />
</form>
</body>
</html>

2.CheckLogin

package edu.cslg.servlet;

import java.io.IOException;
import java.io.PrintWriter;
import javax.servlet.ServletException;
import javax.servlet.annotation.WebServlet;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;

@WebServlet(name = "checkLogin", urlPatterns = { "/checkLogin" })
public class CheckLogin extends HttpServlet {
private static final long serialVersionUID = 1L;
protected void doPost(HttpServletRequest request,
HttpServletResponse response) throws ServletException, IOException {
System.out.println(1);
String loginName = request.getParameter("loginName");
System.out.println(loginName);
PrintWriter out = response.getWriter();
if (loginName.equals("1")) {
out.print("ok");
} else {
out.print("error");
}
}
}

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