AJAX和数据库进行交互判定用户是否注册
- php中世界最好的语言原创
- 2018-04-02 16:35:301389浏览
这次给大家带来AJAX和数据库进行交互判定用户是否注册，AJAX和数据库进行交互判定用户是否注册的注意事项有哪些，下面就是实战案例，一起来看一下。
在很多注册的页面上，我们可能会遇到以下情况，当我们注册用户名的时候可能会提示该用户名已注册，其实现是应用AJAX技术。
首先写一个登陆页面
<html>
<head>
<title></title>
<script type="text/javascript">
var xmlHttp;
var flag;
function createXMLHttp(){
if(window.XMLHttpRequest){
xmlHttp=new XMLHttpRequest();
}else{
xmlHttp=new ActiveXObject("Microsoft.XMLHTTP");
}
}
function checkUserid(userid){
createXMLHttp();
xmlHttp.open("POST","CheckServlet?userid="+userid);
xmlHttp.onreadystatechange=checkUseridCallback;
xmlHttp.send();
document.getElementById("msg").innerHTML="正在验证。。。";
}
function checkUseridCallback(){
if(xmlHttp.readyState==400){
if(xmlHttp.status==200){
var text=xmlHttp.responseText;
if(text=="true"){
flag=false;
document.getElementById("msg").innerHTML="用户ID重复，无法使用";
}else{
flag=true;
document.getElementById("msg").innerHTML="此用户ID可以注册";
}
}
}
}
function checkForm(){
return flag;
}
</script>
</head>
<body>
<form action="tt.jsp" method="post" onsubmit="return checkForm()">
用户ID <input type="text" name="userid" onblur="checkUserid(this.value)"><span id="msg"></span><br>
姓名：<input type="text" name="name"><br>
密码：<input type="password" name="password"><br>
<input type="button" value="注册">
<input type="reset" value="重置">
</form>
</body>
</html>
随后写一个servlet的Java代码
import java.io.IOException;
import java.io.PrintWriter;
import java.sql.Connection;
import java.sql.DriverManager;
import java.sql.PreparedStatement;
import java.sql.ResultSet;
import java.sql.SQLException;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
public class CheckServlet extends HttpServlet{
public static final String DBDRIVER = "oracle.jdbc.OracleDriver";
public static final String DBURL = "jdbc:oracle:thin:@59.173.240.149:1521:heer";
public static final String DBUSER = "hnsyu_dev";
public static final String DBPASS = "hnsyuok";
public void doGet(HttpServletRequest request,HttpServletResponse response)
throws ServletException,IOException{
this.doPost(request, response);
}
public void doPost(HttpServletRequest request,HttpServletResponse response)
throws ServletException,IOException{
request.setCharacterEncoding("gbk");
response.setContentType("text/html");
Connection connection = null;
PreparedStatement preparedStatement = null;
ResultSet resultSet = null;
PrintWriter out = response.getWriter();
String userid = request.getParameter("userid");
try {
Class.forName(DBDRIVER);
connection = DriverManager.getConnection(DBURL, DBUSER, DBPASS);
String sql = "select count(userid) from userdemo where userid=?";
preparedStatement = connection.prepareStatement(sql);
preparedStatement.setString(1,userid);
resultSet = preparedStatement.executeQuery();
if (resultSet.next()) {
if(resultSet.getInt(1)>0){
out.print("false");
System.out.println("true");
}else {
out.print("false");
}
}
out.close();
} catch (Exception e) {
e.printStackTrace();
}finally{
try {
connection.close();
} catch (Exception e) {
e.printStackTrace();
}
}
}
}
其中还需要在web.xml中进行配置
<?xml version="1.0" encoding="UTF-8"?> <web-app version="3.0" xmlns="http://java.sun.com/xml/ns/javaee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_3_0.xsd"> <display-name></display-name> <welcome-file-list> <welcome-file>index.jsp</welcome-file> </welcome-file-list> <servlet> <servlet-name>CheckServlet</servlet-name> <servlet-class>CheckServlet</servlet-class> </servlet> <servlet-mapping> <servlet-name>CheckServlet</servlet-name> <url-pattern>/CheckServlet</url-pattern> </servlet-mapping> </web-app>
相信看了本文案例你已经掌握了方法，更多精彩请关注php中文网其它相关文章！
推荐阅读：
以上是AJAX和数据库进行交互判定用户是否注册的详细内容。更多信息请关注PHP中文网其他相关文章！
声明：
本文内容由网友自发贡献，版权归原作者所有，本站不承担相应法律责任。如您发现有涉嫌抄袭侵权的内容，请联系admin@php.cn