從數據庫my中的username用戶表裡驗證：
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> </p><p>
姓名： <input type="text" name="username" />
</p>
<input type="button" value="check it" onclick="check_username()" />
<p> </p>
<p> </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");
}
?>