從數據庫my中的username用戶表里驗證:
checkusername.html:
復制代碼 代碼如下:
!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
html xmlns="http://www.w3.org/1999/xhtml">
head>
meta http-equiv="Content-Type" content="text/html; charset=gb2312" />
title>無標題文檔/title>
/head>
script language="javascript">
var xmlHttp;
function createXMLHttpRequest(){
if(window.ActiveXObject){
xmlHttp = new ActiveXObject("microsoft.XMLHTTP");
}
else if(window.XMLHttpRequest){
xmlHttp = new XMLHttpRequest();
}
}
function send_request(url,data){
createXMLHttpRequest();
xmlHttp.open("POST",url,true);
xmlHttp.onreadystatechange = check_lll;
xmlHttp.setRequestHeader("CONTENT-TYPE", "application/x-www-form-urlencoded");
xmlHttp.send("username=" + data);
}
function check_lll(){
if(xmlHttp.readyState == 4){
if(xmlHttp.status == 200){
alert(xmlHttp.responseText);
}
}
}
function check_username(){
var f = document.form1;
var username = f.username.value;
if(username == ""){
alert("NULL");
return false;
}
else{
send_request("check_it.php",username);
}
}
/script>
body>
form id="form1" name="form1" method="post" action="">
p>nbsp;/p>p>
姓名: input type="text" name="username" />
/p>
input type="button" value="check it" onclick="check_username()" />
p>nbsp;/p>
p>nbsp; /p>
/form>
/body>
/html>
check_it.php:
復制代碼 代碼如下:
?php
$username = $_POST["username"];
$conn = mysql_connect("localhost:3306","root","123");
mysql_select_db("my",$conn);
$sql = "select * from username where username = '$username'";
$result = mysql_query($sql,$conn);
$num = mysql_fetch_array($result);
if($num > 0){
printf("can't use");
}
else{
printf("It can use");
}
?>
您可能感興趣的文章:- Ajax 用戶名驗證是否存在
- asp.net結合Ajax驗證用戶名是否存在的代碼
- Asp.net下利用Jquery Ajax實現用戶注冊檢測(驗證用戶名是否存)
- jQuery.ajax 用戶登錄驗證代碼
- 基于jQuery實現的Ajax 驗證用戶名是否存在的實現代碼
- Ajax實時驗證用戶名/郵箱等是否已經存在的代碼打包
- PHP+Ajax異步通訊實現用戶名郵箱驗證是否已注冊( 2種方法實現)
- jsp+ajax實現無刷新(鼠標離開文本框即驗證用戶名)實現思路
- Ajax實現異步刷新驗證用戶名是否已存在的具體方法
- Ajax驗證用戶的唯一性