ThinkPHP中ajax使用实例教程，thinkphpajax
ThinkPHP中ajax使用实例教程，thinkphpajax
本文实例讲述了ThinkPHP中使用ajax的方法，提交表单如下图所示：
点击提交，不需要刷新本页，将内容提交到数据库当中，并在本页显示提交的内容。如下图所示：
一、jquery实现方法：
MessageAction.class.php页面代码如下：
<?php
class MessageAction extends Action{
function index(){
$this->display();
}
function add(){
//ajaxReturn(数据,'提示信息',状态)
$m=M('message');
if($m->add($_GET)){
$this->ajaxReturn($_GET,'添加信息成功',1);
}else{
$this->ajaxReturn(0,'添加信息失败',0);
}
}
}
?>
模板index.html代码如下：
<html>
<head>
<script type="text/javascript" src="__PUBLIC__/js/jquery-1.7.1.min.js"></script>
<script type="text/javascript">
$(function(){
$('input:button').click(function(){
var $title=$('input[name="title"]').val();
var $message=$('input[name="message"]').val();
$mess=$('#mess');
$.getJSON('__URL__/add',{title:$title,message:$message},function(json){
//alert(json);return false;
if(json.status==1){
$mess.slideDown(3000,function(){
$mess.css('display','block');
}).html('标题为'+json.data.title+'信息为'+json.data.message);
}else{
$mess.slideDown(3000,function(){
$mess.css('display','block');
}).html('信息添加失败，请检查');
}
});
})
})
</script>
</head>
<body>
<div style="display:none; color:red;" id="mess"></div>
<form action="" method="get">
标题：<input type="text" name="title" /><br />
信息：<input type="text" name="message" /><br />
<input type="button" value="提交" />
</form>
</body>
</html>
二、ThinkPHP实现方法：
MessageAction.class.php页面代码如下：
<?php
class MessageAction extends Action{
function index(){
$this->display();
}
function addtwo(){
$m=M('message');
if($vo=$m->create()){
if($m->add()){
$this->ajaxReturn($vo,'添加成功',1);
}else{
$this->ajaxReturn(0,'添加失败',0);
}
}else{
$this->error($m->getError());
}
}
}
?>
模板index.html代码如下：
<html>
<head>
<script type="text/javascript" src="__PUBLIC__/Js/Base.js"></script>
<script type="text/javascript" src="__PUBLIC__/Js/prototype.js"></script>
<script type="text/javascript" src="__PUBLIC__/Js/mootools.js"></script>
<script type="text/javascript" src="__PUBLIC__/Js/ThinkAjax.js"></script>
<script type="text/javascript">
function add(){
//ThinkAjax.sendForm(表单ID,URL,回调函数,信息显示的地方);
ThinkAjax.sendForm('frm','__URL__/addtwo',wc);
}
function wc(data,status){
if(status!=1){
alert('发送失败');
}else{
$('list').innerHTML+='标题'+data.title+',信息'+data.message;
}
}
</script>
</head>
<body>
<div id="list"></div>
<form action="" method="POST" id="frm">
标题：<input type="text" name="title" /><br />
信息：<input type="text" name="message" /><br />
<input type="button" value="提交" onClick="add()" />
</form>
</body>
</html>
感兴趣的朋友可以测试运行一下本文所示实例，可以加深对Ajax应用的理解。
thinkphp自带有ajax方法，很简单的使用，方便的交互。举一个例子你就明白啦。
先包含所需要的js(他的官网里有下载)，路径自己修改。
你那个回调函数里的 参数写错
这样写
alert(dd.data);
你没了解 ajaxReturn返回的数据格式
你可以直接通过 浏览器访问 is_user 就可以看到
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