AJAX PHP无刷新form表单提交的简单实现(推荐)

ajax.php：

<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.getElementByIdx_x("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;
  }
}
alert (userName);
}
</script>
<body >
<div id="msg"></div>
<form name="user_info" method="post" action="">
姓名：<input type="text" id="user_name"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>

ajax_output.php：

<?php

   $username = $_POST['user_name'];
   $userage = $_POST['user_age'];
   $usersex = $_POST['user_sex'];
  echo "$username <br>";
  echo "$userage <br>";
  echo "$usersex <br>";

  $db = new mysqli('localhost','root','123456','test');
  if(!$db){
  echo "连接失败！";
  }
  $db->query("set names utf8");
  $query = "insert into userinfo(uname,uage,usex) values ('".$username."','".$userage."','".$usersex."')";
  $result = $db->query($query);
  if ($result){
  echo "上传成功！";
  }
  else {
  echo "失败！";
  }
  $db->close();

?>

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AJAX PHP无刷新form表单提交的简单实现(推荐)

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