ajax php用户无刷新登录实例
<!doctype html>
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="content-type" content="text/html; charset=gb2312" />
<title>ajax php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
<?php
$uid = $_post['uname'];
$pwd = $_post['upwd'];
$sql = "select * from tabname where uid='$uid' and pwd='$pwd'";
$query = mysql_query($sql);
if (mysql_num_rows($query)) {
echo '登录成功';
} else {
echo '用户名或密码不正确!';
}
?>教程地址：http://www.phprm.com/develop/fs976.html
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