Extjs4创建简单的用户登录

首先给出Extjs代码:

Ext.define('MyApp.view.ui.MyForm', {
extend: 'Ext.form.Panel',
height: 124,
width: 343,
bodyPadding: 10,
title: '用户登录',
initComponent: function() {
var me = this;
me.items = [
{
xtype: 'textfield',
name: 'username',
fieldLabel: '用户名',
labelWidth: 60,
allowBlank: false,
anchor: '100%',
emptyText:'请填写用户名'
},
{
xtype: 'textfield',
name: 'password',
inputType:'password',
fieldLabel: '密码',
labelWidth: 60,
allowBlank: false,
anchor: '100%',
emptyText:'请填写密码'
},
{
xtype: 'container',
padding: '0 0 0 65',
items: [
{
xtype: 'button',
text: '登录',
handler:function(btn){
var form=btn.up('form').getForm();
if(form.isValid())
{
form.submit({
url:'pro.php',
success:function(){
location.href="main.php";
},
failure:function(grid,action)
{
Ext.Msg.alert('信息提示',action.result.msg);
}

});
}
else
{
Ext.Msg.alert('信息提示','请填写用户名和密码');
}

}
},
{
xtype: 'button',
margin: '0 0 0 10',
text: '重置',
handler:function(btn){
btn.up('form').getForm().reset();
}
}
]
}
];
me.callParent(arguments);
}
});


Ext.onReady(function(){
Ext.create('MyApp.view.ui.MyForm',{
renderTo:Ext.get('main')
});
});

下面给出服务端代码(PHP):

<?php

$username=$_POST['username'];
$password=$_POST['password'];

if($username=='root' && $password=='123')
{
echo "{
success:true,
msg:'登陆成功!'
}";

}
else
{
echo "{
success:false,
msg:'登录失败!'
}";
}

?>



posted on 2012-03-28 19:31  crazymus  阅读(443)  评论(0编辑  收藏  举报

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