AJAX PHP无刷新form表单提交的简单实现(推荐)_PHP教程

AJAX PHP无刷新form表单提交的简单实现(推荐)

2021-02-27 16:51PHP教程网 PHP教程

下面小编就为大家带来一篇AJAX PHP无刷新form表单提交的简单实现(推荐)。小编觉得挺不错的，现在就分享给大家，也给大家做个参考。一起跟随小编过来看看吧

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();

									?>