求ajax代码，后台php
form表单内容 name ,age,sex，提交到后台index.php的insert()方法，用post方式。求详细代码，实在是不会呀
回复讨论(解决方案)
建议 LZ 还是仔细学习一下前端的知识
sample.html
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"><html> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>HTML</title> <script src="jquery.js"type="text/javascript"></script> <script type="text/javascript"> var data={name:'orange',age:20,sex:'male'}; var url='index.php'; jQuery(document).ready(function(){ jQuery('#submit').click( function(){ jQuery.post(url,data,function(result){ if(result.code==0){ alert('插入成功'); }else{ alert('插入失败，错误信息:'+result.msg); } },'json').error(function(){ alert("网络不通或者服务器执行失败"); }); } ); }); </script> </head> <body> <input type="button" id="submit" value="提交" /> </body></html>index.php
<?php function insert(){ $ret = array('code'=>1,'msg'=>'数据不合法'); extract($_POST); /* * 执行逻辑 * 插入数据库 $name $age $sex */ $ret['code'] = 1; echo json_encode($ret); } insert(); //表单代码
<form action="index.php" method="post"> 姓名：<input type="text" name="name" /><br /> 年龄：<input type="text" name="age" /><br /> 性别：<input type="text" name="sex" /><br /> <input type="submit" value="提交" /></form>
//index.php代码
$name=$_POST['name'];$age=$_POST['age'];$sex=$_POST['sex']; function insert($name,$age,$sex){ echo $name.$age.$sex; }insert($name,$age,$sex);好吧，有这么多，我就不提交了。。。多学习吧。网上很多资料
我也在学JQ，用JQ得到要提交的值，在拼装。提交到服务端处理和普通程序一样。
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