日期:2014-05-16  浏览次数:20787 次

php、Ajax、Post的传值的例子,运行有问题,帮忙看看!!!!!!
这是我在网上找到的一个例子,我运行时有问题,帮忙看看!http://cjmxp007.blog.163.com/blog/static/3547383720074252134287/
JScript code
php Ajax Post 传值

用Ajax 进行Post传值

以下程序已调试通过:

<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>


<script language="javascript">
function saveUserInfo()
{
//获取接受返回信息层
var msg = document.getElementById("msg");

//获取表单对象和用户信息值
var f = document.user_info;
var userName = f.user_name.value;
var userAge   = f.user_age.value;
var userSex   = f.user_sex.value;

//接收表单的URL地址
var url = "/ajax_output.php";

//需要POST的值,把每个变量都通过&来联接
var postStr   = "user_name="+ userName +"&user_age="+ userAge +"&user_sex="+ userSex;

//实例化Ajax
//var ajax = InitAjax();

          var ajax = false;
         //开始初始化XMLHttpRequest对象
         if(window.XMLHttpRequest) { //Mozilla 浏览器
                 ajax = new XMLHttpRequest();
                 if (ajax.overrideMimeType) {//设置MiME类别
                         ajax.overrideMimeType("text/xml");
                 }
         }
         else if (window.ActiveXObject) { // IE浏览器
                 try {
                         ajax = new ActiveXObject("Msxml2.XMLHTTP");
                 } catch (e) {
                         try {
                                 ajax = new ActiveXObject("Microsoft.XMLHTTP");
                         } catch (e) {}
                 }
         }
         if (!ajax) { // 异常,创建对象实例失败
                 window.alert("不能创建XMLHttpRequest对象实例.");
                 return false;
         }
                
                
                

//通过Post方式打开连接
ajax.open("POST", url, true);

//定义传输的文件HTTP头信息
ajax.setRequestHeader("Content-Type","application/x-www-form-urlencoded");

//发送POST数据
ajax.send(postStr);

//获取执行状态
ajax.onreadystatechange = function() { 
   //如果执行状态成功,那么就把返回信息写到指定的层里
   if (ajax.readyState == 4 && ajax.status == 200) { 
    msg.innerHTML = ajax.responseText; 
   } 
} 
}
</script>
<body >
<div id="msg"></div>
<form name="user_info" method="post" action="">
姓名:<input type="text" name="user_name" /><br />
年龄:<input type="text" name="user_age" /><br />
性别:<input type="text" name="user_sex" /><br />

<input type="button" value="提交表单" onClick="saveUserInfo()">
</form>

</body>


以上页面存为ajax.php

然后再建 一个PHP文件,ajax_output.php

<?
     echo $_POST['user_name'];
     echo $_POST['user_age'];
     echo $_POST['user_sex'];
?>




------解决方案--------------------
这样试试:
HTML code
<script language="javascript">
function saveUserInfo(f){
  var userName = f.user_name.value;
  var userAge   = f.user_age.value;
  var userSex   = f.user_sex.value;

  var url = "/ajax_output.php?"+new Date().getTime=

------解决方案--------------------
ajax_output.php
PHP code

<?
  if(empty($_POST['username'])){die('ajax failed!')}
  echo "ajax success!<br>";
  echo $_POST['username'];
  echo $_POST['userage'];
  echo $_POST['usersex'];
?>

------解决方案--------------------
var url = "/ajax_output.php";
不用/吧,
var userName = f.user_name.value;
var userAge = f.user_age.value;
var userSex = f.user_sex.value;
取值尽量用documen.getElementById();为每一个input加个id属性


if (ajax.readyState == 4){