Javascript 如何使用AJAX（不是Jquery AJAX）发送参数并使用PHP接收参数

我正在尝试使用HTML表单值向服务器发送一些AJAX（不是jQueryAjax）参数，并将其存储到数据库中。但它不能正常工作。当我按下“让我们开始”按钮时，我收到了错误消息 这是我的密码

<div class ="col-xs-12 col-md-12 col-sm-12 col-lg-12" id ="AJAXsignup">
        <h3>Creating a new account...</h3>

        <input class ="form-control" type ="text" name ="fullname" placeholder ="Type your name eg:-Jhon Smith" id ="namefull">
        <span class ="label label-warning" id ="err_fullname">You should type your name</span>

        <input class ="form-control" type ="text" name ="username" placeholder ="Type your user name eg:-Jhon 95" id ="nameuser" style ="margin-top:5px;">
        <span class ="label label-warning" id ="err_username">Missing username</span>

        <div class ="input-group">
        <span class ="input-group-addon">@</span>
        <input class ="form-control" type ="email" name ="email" placeholder ="Type your Email" id ="email" style ="margin-top:5px;">
        <span class ="label label-warning" id ="err_email">Missing email</span>
        </div>

        <input class ="form-control" type ="password" name ="password" placeholder ="Type your Password" id ="pass" style ="margin-top:5px;">
        <span class ="label label-warning" id ="err_password">Missing Password</span>

        <input class ="form-control" type ="password" name ="password" placeholder ="Repeat your Password" id ="repass" style ="margin-top:5px;">
        <span style ="margin-bottom:5px;" class ="label label-warning" id ="err_repass">Repeat Password</span>

        <img class ="col-sm-offset-5 col-md-offset-5 col-lg-offset-5" src ="images\Preloader_2.gif" id ="loader" style ="display:none;" style ="margin-bottom:5px;">

        <input class ="btn btn-lg btn-info col-md-12 col-lg-12 col-sm-12 col-xs-12" type ="submit" id ="btn" value ="Let's start!" style ="margin-bottom:5px;" onclick = "validate()">

        <input class ="btn btn-lg btn-danger col-md-12 col-lg-12 col-sm-12 col-xs-12"type ="button" id ="frm-can" value ="No.Thanks">

    </div>

PHP代码

$fullname = $_GET['w'];
$username = $_GET['n_p'];
$email = $_GET['tv'];
$password = $_GET['q'];
$server = "localhost";
$username = "root";
try{
    $conn = new PDO("mysql:host=$server;dbname=reg_mem",$username);
    $conn->setAttribute(PDO::ATTR_ERRMODE,PDO::ERRMODE_EXCEPTION);
    $sql = "INSERT INTO mem_info(Full_name,User_name,Email,Password) VALUES($fullname,$username,$email,$password)";
    $conn->exec($sql);
    echo "The infomation sent sunncessfully.";
    }catch(PDOException $e){
    echo "The infomation unable to send right now";
    }

---

如果要使用GET方法发送信息，请将信息添加到URL：

xhttp.open("GET", "filename.php?fname=value&lname=value", true);
xhttp.send();

要像HTML表单一样发布数据，请使用setRequestHeader（）添加HTTP标头。在send（）方法中指定要发送的数据：

---

使用Jquery ajax将var发送到PHP页面

$.ajax({
        type: "GET", (or POST)
        url: "code.php",
        data: { fullname, user_n, email },
        cache: false,
        success: function(){
      }
    });

});

---

只需从HTML更改这一行

        <input class ="btn btn-lg btn-info col-md-12 col-lg-12 col-sm-12 col-xs-12" type ="submit" id ="btn" value ="Let's start!" style ="margin-bottom:5px;" onclick = "validate()">

到

您只需将您的

onclick=“validate（）”

更改为

onclick=“sendingformation（）；”

或者您可以在单击时调用2个函数，比如

onclick=“sendingformation（）；validate（）；“

假设您的

validate（）

工作正常并且

发送信息（）

是从

validate（）

函数中调用的。我制作了代码的工作副本。我在

$（“#loader”）中发现了一个错误。如果没有包含
jQuery
，则不能调用
fadeOut（）
。但是如果您已经包含了
jQuery
，那么就不需要javascript函数，您可以使用jqueryajax本身

除此之外，您的代码工作正常。
您遇到了哪个错误？
$.ajax({
        type: "GET", (or POST)
        url: "code.php",
        data: { fullname, user_n, email },
        cache: false,
        success: function(){
      }
    });

});

        <input class ="btn btn-lg btn-info col-md-12 col-lg-12 col-sm-12 col-xs-12" type ="submit" id ="btn" value ="Let's start!" style ="margin-bottom:5px;" onclick = "validate()">

        <input class ="btn btn-lg btn-info col-md-12 col-lg-12 col-sm-12 col-xs-12" type ="submit" id ="btn" value ="Let's start!" style ="margin-bottom:5px;" onclick = "sendinginfomation()">