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ajax php用户无刷新登录实例

WBOY
Release： 2016-05-25 16:51:17
Original
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<!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(&#39;content-type&#39;, &#39;application/x-www-form-urlencoded&#39;);
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>
Copy after login

userlogin.php文件

<?php
$uid = $_post[&#39;uname&#39;];
$pwd = $_post[&#39;upwd&#39;];
$sql = "select * from tabname where uid=&#39;$uid&#39; and pwd=&#39;$pwd&#39;";
$query = mysql_query($sql);
if (mysql_num_rows($query)) {
    echo &#39;登录成功&#39;;
} else {
    echo &#39;用户名或密码不正确!&#39;;
}
?>
Copy after login


教程地址：

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Related labels：
ajax Example No refresh user Log in
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