Question

我的问题的简短版本：我点击“登录”弹出一个登录表单。然后用户可以通过输入电子邮件和密码登录。提交表单并成功后，用户将重定向到他的个人资料。

我想要的：如果验证中出现错误，例如无效的电子邮件或密码错误，则不应提交登录表单，而是在登录弹出窗口中显示错误消息。因此，如果出错，弹出窗口应保持打开状态。

代码：

弹出登录表单：

<div id="added_login">
    <div class="form_title">
        <h3>Anmelden</h3>
        <img id="form_close" src="images/close.png"/>
    </div>
    <form id="login_form" action="includes/login.inc.php" method="POST">
        <img src="images/or_line.png" />
        <p>Melde dich mit deiner E-Mail Adresse an</p>
        <input id="email" type="text" name="email" placeholder="E-Mail Adresse oder Benutzername">
        <input id="password" type="password" name="password" placeholder="Passwort eingeben">
        <div class="error_msg_container"></div>
        <input id="submit" class="button button_highlight" type="submit" name="submit_login" value="Anmelden">
        <div class="login_footer">
            <p class="forget_password">Passwort vergessen?</p>
            <p id="loginform_register" class="button_register">Registrieren</p>
        </div>
    </form>
</div>

弹出窗体的jquery / ajax部分

$("#login_form").submit(function(e){
    e.preventDefault();
    var email = $("#email").val();
    var password = $("#password").val();
    var submit = $("#submit").val();
    $.ajax({
        type: "POST",
        url: "includes/login.inc.php",
        data: {
            email: email,
            password: password,
            submit_login: submit
        },
        success: function(data){
            $("#error_msg_container").html(data);
        },
        error: function(){
            alert("ERROR");
        }
    })
});

数据库的php部分：

<?php

session_start();

//this function checks the user input
function IsEmail($email){
    if(filter_var($email, FILTER_VALIDATE_EMAIL)){
        //if user input is an email then return true
        return true;
    }else{
        //if user input isnt an email then return false
        return false;
    }
}

if (isset($_POST['submit_login'])) {
    include_once 'db_connect.inc.php';

    $loginid = mysqli_real_escape_string($conn, $_POST['email']);
    $password = mysqli_real_escape_string($conn, $_POST['password']);

    if(empty($loginid)){
        //Check for empty email
        echo "<p>E-Mail cannot be empty.</p>";
    }elseif(empty($password)){
        //Check for empty password
        echo "<p>Password cannot be empty.</p>";
    }else{
        $check_email = IsEmail($loginid);
        if($check_email){
            //is input an email
            $sql = "SELECT * FROM users WHERE email='$loginid'";
            $result = mysqli_query($conn, $sql);
        }else{
            //is email a username
            $sql = "SELECT * FROM users WHERE username='$loginid'";
            $result = mysqli_query($conn, $sql);
        }
        $resultCheck = mysqli_num_rows($result);
        //Check for user existing
        if($resultCheck < 1){
            echo "<p>No such a user found.</p>";
        }else{
            if($row = mysqli_fetch_assoc($result)){
                //Dehashing password
                $hashedPasswordCheck = password_verify($password, $row['password']);
                //Check for correct password
                if($hashedPasswordCheck == false){
                    header("Location: ../index.php?login=wrong_password");
                    exit();
                }elseif($hashedPasswordCheck == true){

                    if($row[active] == 0){
                        echo "<p>Your accent have to set on active</p>";
                    }else{
                        //Log in the user
                        $_SESSION[user_id] = $row[id];
                        $_SESSION[user_firstname] = $row[firstname];
                        $_SESSION[user_lastname] = $row[lastname];
                        $_SESSION[user_gender] = $row[gender];
                        $_SESSION[user_username] = $row[username];
                        $_SESSION[user_email] = $row[email];
                        $_SESSION[user_level] = $row[level];
                        $_SESSION[user_points] = $row[points];
                        $_SESSION[loggedIn] = true;

                        header("Location: ../index.php?login=success");
                        exit();
                    }
                }
            }
        }
    }

}else{
    header("Location: ../index.php");
    exit();
}

现在的问题是什么都没发生..弹出登录窗口保持打开但错误信息没有显示，即使输入正确也没有发生任何事情..

希望你们能帮助我。 THX

Answer 1

There is issue in your jquery code.

在ajax成功函数中，你提到了div id＆＃39; error_msg_container＆＃39;这实际上是html中的一个类。代码应该是

    success: function(data){
        $(".error_msg_container").html(data);
    }

通过验证错误

1 个答案: