AJAX PHP无刷新form表单提交的简单实现(推荐) - 哈喽比特

1359次阅读 | 发布于5年以前

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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