很菜的前后台交互程序，用jq,ajax，json格式，解决方案

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Release： 2016-06-13 12:11:46

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很菜的前后台交互程序，用jq,ajax，json格式，
前台输意一个数字，后台处理，比如加2，再返回前台，求教。请别嘲笑。
------解决思路----------------------
html代码部分：

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"><br /><html xmlns="http://www.w3.org/1999/xhtml"><br /><head><br /><meta http-equiv="Content-Type" content="text/html; charset=utf-8" /><br /><title>无标题文档</title><br /><script src="http://cloud.xing-xing.com/jquery.js"></script><br /><script type="text/javascript" language="javascript"><br />$(function(){<br />	$.get('handle.php',{num:1},function(data){<br />		$("div").html(data);//在div里面输出3<br />	})	<br />})<br /></script><br /></head><br /><body><br /><div></div><br /></body><br /></html>

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handle.php代码部分：

<?php<br />echo $_GET['num']+2;<br />?>

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