ajax php用户无刷新登录实例
<!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>ajax php教程用户无刷新登录实例</title>
<script>
function userlogin(){
var xmlhttp;
var str;
var sendstr="";
try{
xmlhttp=new xmlhttprequest();
}
catch(e){
xmlhttp=new activexobject("microsoft.xmlhttp");
}
xmlhttp.onreadystatechange=function(){
if (xmlhttp.readystate==4){
if (xmlhttp.status==200){
str = xmlhttp.responsetext;
document.getelementbyid("userlogin").innerhtml=str;
}else{
alert("系统错误,如有疑问,请与管理员联系!"+xmlhttp.status);
}
}
}
xmlhttp.open("post","config/userlogin.php",true);
xmlhttp.setrequestheader('content-type','application/x-www-form-urlencoded');
xmlhttp.send(sendstr);
}
</script>
</head><body>
<form id="form1" name="form1" method="post" action="">
<p>
<label for="textfield"></label>
<input type="text" name="uname" id="uname" /><span id="userlogin"></span><br />
<input type="text" name="upwd" id="upwd" /><span id="upwds"></span>
输入用户名</p>
<p>
<input type="button" name="button" id="button" value="登录" onclick="userlogin();" />
</p>
</form>
</body>
</html>
userlogin.php文件
<?
$uid = $_post['uname'];
$pwd = $_post['upwd'];
$sql ="select * from tabname where uid='$uid' and pwd='$pwd'";
$query = mysql教程_query( $sql );
if( mysql_num_rows( $query ) )
{
echo '登录成功';
}
else
{
echo '用户名或密码不正确!';
}
?>
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