用Ajax 進行Post傳值
以下程序已調試通過:
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<script language="javascript">
function saveUserInfo()
{
//獲取接受返回信息層
var msg = document.getElementByIdx_x("msg");
//獲取表單對象和用戶信息值
var f = document.user_info;
var userName = f.user_name.value;
var userAge
= f.user_age.value;
var userSex
= f.user_sex.value;
//接收表單的URL地址
var url = "/ajax_output.php";
//需要POST的值,把每個變量都通過&來聯接
var postStr
= "user_name="+ userName +"&user_age="+ userAge +"&user_sex="+ userSex;
//實例化Ajax
//var ajax = InitAjax();
//通過Post方式打開連接
ajax.open("POST", url, true);
//定義傳輸的文件HTTP頭信息
ajax.setRequestHeader("Content-Type","application/x-www-form-urlencoded");
//發送POST數據
ajax.send(postStr);
//獲取執行狀態
ajax.onreadystatechange = function() {
}
}
</script>
<body >
<div id="msg"></div>
<form name="user_info" method="post" action="">
姓名:<input type="text" name="user_name" /><br />
年齡:<input type="text" name="user_age" /><br />
性別:<input type="text" name="user_sex" /><br />
<input type="button" value="提交表單" onClick="saveUserInfo()">
</form>
</body>
以上頁面存為ajax.php
然后再建 一個PHP文件,ajax_output.php
<?
?>