Javascript Ajax和php，生成一个div

我想在“result”div中显示服务器页面的结果。 这是我的计划：

home.php

 <script src="myscripts.js"></script>
 <div id="loginform">
   <div id="result"></div>
   <?php include_once 'core.login.php'; ?>
 </div>

<?php

 // stuff about $db variable e other stupid things like $tool = new Tools etc...

 $email = $_POST["email"];
 $password = $_POST["password"];

// Initializing login process
if (isset($email) or isset($password)) {
$tool->decode($email, $password, $db);
}

?>

 <form action="core.login.php" method="POST" id="login">
        <fieldset>
            <legend>Data login:</legend>
            Email:<input id="email" type="email" name="email" placeholder="someone@example.com" required>
            <br>
            Password:<input id="password" type="password" name="password" required>
            <br>
            <input type="submit" value="Submit">
        </fieldset>
 </form>

public function decode($email, $password, $db) {
    if ($stmt = $db->prepare("SELECT password FROM users WHERE email = ?")) {
        $stmt->bind_param('s', $email);
        $stmt->execute();
        $stmt->store_result();
        $stmt->bind_result($password_db);
        $stmt->fetch();
        if ($stmt->num_rows == 1) {
            if ($password == $password_db) {
                echo "Success"; 
            } else {
                echo "Error";      
            }
        } else {
            echo "Email didn't found!";
        }
    }
}

core.login.php

 <script src="myscripts.js"></script>
 <div id="loginform">
   <div id="result"></div>
   <?php include_once 'core.login.php'; ?>
 </div>

<?php

 // stuff about $db variable e other stupid things like $tool = new Tools etc...

 $email = $_POST["email"];
 $password = $_POST["password"];

// Initializing login process
if (isset($email) or isset($password)) {
$tool->decode($email, $password, $db);
}

?>

 <form action="core.login.php" method="POST" id="login">
        <fieldset>
            <legend>Data login:</legend>
            Email:<input id="email" type="email" name="email" placeholder="someone@example.com" required>
            <br>
            Password:<input id="password" type="password" name="password" required>
            <br>
            <input type="submit" value="Submit">
        </fieldset>
 </form>

public function decode($email, $password, $db) {
    if ($stmt = $db->prepare("SELECT password FROM users WHERE email = ?")) {
        $stmt->bind_param('s', $email);
        $stmt->execute();
        $stmt->store_result();
        $stmt->bind_result($password_db);
        $stmt->fetch();
        if ($stmt->num_rows == 1) {
            if ($password == $password_db) {
                echo "Success"; 
            } else {
                echo "Error";      
            }
        } else {
            echo "Email didn't found!";
        }
    }
}

没有AJAX如果我正常启动代码，它会给我正确的响应结果（成功或错误），但是当我使用AJAX时，什么也不会发生

更新

好了，伙计们，问题是操作url，我的核心文件在core/core.login.php文件夹中，现在它在result div中显示页面，但是页面核心现在显示以下内容： 致命错误：对第9行网站中的非对象调用成员函数decode（）

---

也许ajax没有像object那样传递变量？

请在js文件中尝试：

 (Or Document)
$('#loginform').on('submit', '#login', function() {
  //AJAX etc..
})

而不是：

$('#login').submit(function() {
       //AJAX etc..
}

试试这个js

$(document).ready(function() {
$('#login').submit(function() { // catch the form's submit event
$.ajax({ // create an AJAX call...
data: $("#login").serialize(), // get the form data
type: $(#login).attr('method'), // GET or POST
url: $(#login).attr('action'), // the file to call
success: function(response) { // on success..
    $('#result').html(response); // update the DIV
}
});
 return false; // cancel original event to prevent form submitting
   });
 });

---

你检查控制台有没有错误？观察控制台时，在请求/响应过程中会发生什么。将

error

处理程序添加到ajax调用中，查看错误消息是什么。我认为不会发生任何事情。控制台可能在说些什么。我不太了解ajax的家伙们，我如何检查控制台的错误呢^^“他们正在讨论浏览器中的开发人员控制台。您是否从类中将$tool设置为类。php？？？

$tool=new class（）

错误是：在第8行对网站中的非对象调用成员函数decode（）