php ajax post 登陆ajax请求

<!-- <meta charset="utf-8">
<div style="overflow: hidden; 30px;height: 20px;"> 看不见的部分</div> -->
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<title>jquery控制是否显示</title>
<script type="text/javascript" src="http://cdn.bootcss.com/jquery/2.1.4/jquery.min.js"></script>
<script type="text/javascript">
$(document).ready(function(){
$("#hide").click(function(){
$("#p1").hide();
$("#p2").show();
});
$("#show").click(function(){
$("#p1").show();
$("#p2").hide();
});
});

</script>
</head>
<body>
<p id="p1">显示第一段内容</p>
<p id="p2">显示的第二段内容</p>
<!-- <button id="hide" type="button">隐藏</button>
<button id="show" type="button">显示</button> -->
<input type="radio" id="hide" name="gender">男
<input type="radio" id="show" name="gender">女
</body>
</html>

<?php

header("Content-Type:text/html; charset=utf-8");
$account = $_POST['account'];
$password = $_POST['password'];
// 判断是否正确
$msg['code'] = '200';
$msg['status'] = 'success';
$msg['msg'] = 'ok';
echo json_encode($msg);

原文地址:https://www.cnblogs.com/ayanboke/p/11400529.html