AJAX PHP无刷新form表单提交的简单实现(推荐)

ajax.PHP：

PHP;"> Untitled Document

//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);
}

ajax_output.PHP：

PHP;"> PHP

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

$db = new MysqLi('localhost','root','123456','test');
if(!$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();

以上这篇AJAX PHP无刷新form表单提交的简单实现(推荐)就是小编分享给大家的全部内容了，希望能给大家一个参考，也希望大家多多支持编程之家。

ajax提交form表单 php

AJAX PHP无刷新form表单提交的简单实现(推荐)

相关文章