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AJAX PHP无刷新form表单提交的简单实现(推荐)

2024-05-04 22:52:54
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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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