首页 > 代码库 > AJAX PHP无刷新form表单提交的简单实现(推荐)

AJAX PHP无刷新form表单提交的简单实现(推荐)

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=http://www.mamicode.com/"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(); ?>

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轩宇网工作室。

AJAX PHP无刷新form表单提交的简单实现(推荐)