今天在网上看到一个利用extjs制作ajax登录的页面的test.Html代码，这里和大家分享一下：
test.html代码
<!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=utf-8" />
<title>Extjs 表单</title>
<link type="text/Css" rel="stylesheet" href="/js/ext3.1/resources/css/ext-all.css" />
<script type="text/Javascript" src="/js/ext3.1/adapter/ext/ext-base.js"></script>
<script type="text/JAVAscript" src="/js/ext3.1/ext-all.js"></script>
<script type="text/javascript" src="/js/ext3.1/ext-lang-zh_CN.js"></script>
<script type="text/javascript">
Ext.onReady(function () {
var f = new Ext.form.FormPanel({
title : '欢迎登录',
width : 300,
height : 150,
bodyStyle : 'padding:20px',
labelAlign : 'right',
frame : true,
items : [
new Ext.form.TextField({
name : 'username',
allowBlank : false,
fieldLabel : '用户名'
}),{
name : 'passWord',
xtype : 'textfield',
inputType : 'password',
fieldLabel : '密码',
allowBlank : false
}
],
buttons : [{
text : '确定',
handler : function () {
var username = f.getForm().findField('username').getValue();
var password = f.getForm().findField('password').getValue();
Ext.Ajax.request({
url : 'ajax/ajaxLogin.PHP',
success : function (response,config) {
var flog = response.responseText;
if(0 == flog) {
Ext.MessageBox.alert('提示','登录失败，没有查找到此用户');
} else if(1 == flog) {
Ext.MessageBox.alert('提示','登录失败,密码错误！');
} else {
window.location = 'http://www.test.com';
}
},
failure : function () {
Ext.MessageBox.alert('错误','连接错误！');
},
method : 'post',
params : {username : username,password : password}
});
}
},{
text : '重置',
handler : function () {
f.getForm().reset();
}
}]
})
f.render('a');
});
</script>
</head>
<body>
<div id="a"></div>
</body>
</html>
ajaxLogin.php源代码:
<?php
header("Context-type:text/html;charset=utf-8");
$username = $_REQUEST['username'];
$password = $_REQUEST['password'];
$conn = MySQL_connect('localhost','root',''); mysql_select_db('test'); $sql = "SELECT * FROM user_group WHERE username = '$username'"; $res = mysql_query($sql); while($row = mysql_fetch_array($res)) { if($password == $row['password']) { echo 10;exit; } else { echo 1;exit; } } echo 0; exit;
