我的网站标题中有一个登录模式，当用户单击登录按钮时，该模式会显示在屏幕上。每当我提交登录表单时，它都会创建一个新页面，其中包含错误。相反，我正在寻找登录模式中出现的错误。有人知道我如何实现这一目标吗？
这是登录.php
<?php //checks pword and email match on database
// Check if the form has been submitted:
if ($_SERVER['REQUEST_METHOD'] == 'POST') {
// Requires two additional files
require ('includes/login_functions.inc.php');
require ('C:\xampp\htdocs\final\includes\db.php');
// Check the login.
list ($check, $data) = check_login($dbc, $_POST['email'], $_POST['pass']);
if ($check) { // OK!
// Set the session data.
session_start();
$_SESSION['id'] = $data['id'];
$_SESSION['first_name'] = $data['first_name'];
// Redirect the customer to loggedin page
redirect_user('loggedin.php');
} else { // If unsuccessful
// Assign $data to $errors for login_page.inc.php:
$errors = $data;
}
mysqli_close($dbc); // Close the database connection.
}
// Create the page:
include ('includes/login_page.inc.php');
?>
这是login_page.inc
<link href="css/mystyle.css" rel="stylesheet">
<?php //displays errors and creates form
// Include the header:
$page_title = 'Login';
if (isset($errors) && !empty($errors)) {
echo '<h1><font color="orange">Error!</font> </h1>
<p class="error">The following error(s) occurred:<br />';
foreach ($errors as $msg) {
echo " - $msg<br />\n";
}
echo '</p><p>Please try again.</p>';
}
// Display the form
?>
<div class="modal fade" id="login-modal" tabindex="-1" role="dialog" aria-labelledby="myModalLabel" aria-hidden="true" style="display: none;">
<div class="modal-dialog">
<div class="loginmodal-container">
<h1>Login to Your Account</h1><br>
<form action="login.php" method="post">
<input type="text" name="email" placeholder="Email Address">
<input type="password" name="pass" placeholder="Password">
<input type="submit" name="submit" class="login loginmodal-submit" value="Login">
</form>
<div class="login-help">
<a>Dont have an account? - </a><a href="register.php">Register</a>
</div>
<?php
// Display error messages if there are any
/* if (isset($errors) && !empty($errors)) {
echo '<h1><font color="orange">Error!</font> </h1>
<p class="error">The following error(s) occurred:<br />';
foreach ($errors as $msg) {
echo " - $msg<br />\n";
}
echo '</p><p>Please try again.</p>';
}
*/
?>
</div>
</div>
</div>
这是login_functions.inc
<?php //checks for errors
function redirect_user ($page = 'index.php') {
$url = 'http://' . $_SERVER['HTTP_HOST'] . dirname($_SERVER['PHP_SELF']);
$url = rtrim($url, '/\\');
$url .= '/' . $page;
header("Location: $url");
exit(); // Quit the script.
}
// Checks the validation of the login.
function check_login($dbc, $email = '', $password = '') {
$errors = array(); // Initialize error array.
// Validate the email address:
if (empty($email)) {
$errors[] = 'You forgot to enter your email address.';
} else {
$e = mysqli_real_escape_string($dbc, trim($email));
}
// Validate the password:
if (empty($password)) {
$errors[] = 'You forgot to enter your password.';
} else {
$p = mysqli_real_escape_string($dbc, trim($password));
}
if (empty($errors)) { // If everything's OK.
// Retrieve the customer_id and first_name for that email/password combination:
$q = "SELECT id, first_name FROM user WHERE email='$e' AND password=SHA1('$p')";
$r = @mysqli_query ($dbc, $q); // Run the query.
// Check the result:
if (mysqli_num_rows($r) == 1) {
// Fetch the record:
$row = mysqli_fetch_array ($r, MYSQLI_ASSOC);
// Return true and the record:
return array(true, $row);
} else { // Not a match!
$errors[] = 'The email address and password entered do not match those on file.';
}
} // End of empty($errors) IF.
// Return false and the errors:
return array(false, $errors);
} // End of check_login() function.
最佳答案
这是一个 AJAX 解决方案。对您的文件进行了许多更改，但评论良好。
在login_page.inc中，只有表单受到影响:
<form action="login.php" method="post">
<input type="text" name="email" placeholder="Email Address" id="email">
<input type="password" name="pass" placeholder="Password" id="password">
<input type="submit" name="submit" class="login loginmodal-submit" value="Login">
<div id="login_msg"></div>
然后将整个 login.php 替换为以下代码:
// Check if the form has been submitted:
if (isset($_POST['login'])) {
// Requires two additional files
require ('includes/login_functions.inc.php');
require ('C:\xampp\htdocs\final\includes\db.php');
// Check the login.
list ($check, $data) = check_login($dbc, $_POST['email'], $_POST['pass']); //you'll need to validate your email and password before inputing them into this method
if ($check) { // OK!
// Set the session data.
session_start(); //this should be the first thing in your page, if possible in line 1
$_SESSION['id'] = $data['id'];
$_SESSION['first_name'] = $data['first_name'];
die("true"); //stops here and send response to the AJAX Script
//redirect_user('loggedin.php');
} else { // If unsuccessful
// Assign $data to $errors for login_page.inc.php:
$errors = $data;
}
mysqli_close($dbc); // Close the database connection.
}
// Create the page:
include ('includes/login_page.inc.php');
}
将 JQuery 包含到您的项目中。创建一个名为 ajax.js 的新文件，并将其也链接到您的项目。这是ajax.js的内容:
/*
* Login User
*/
$(".loginmodal-submit").click(function (e) {
e.preventDefault();
var email = $("#email").val();
var password = $("#password").val();
var login = $(".loginmodal-submit").val();
if (email != "" && password != "") {
$("login_msg").html("<i class='fa fa-spinner fa-spin fa-lg text-success' style='font-size:20px'></i>");
$.ajax({
type: 'POST',
url: "login.php",
data: {login: login, email: email, password: password},
success: function (response)
{
if (response == "true") {
$(".login-modal").html("<span class='alert alert-success'>Login successful! Redirecting...</span>");
window.location.href = "loggedin.php";
}
else {
$("#login_msg").html('Sorry, your registration was not successful. Please check your details and try again: ');
console.log(response)
}
}
});
}
});
关于php - 如何将错误消息输入到登录模式中，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/36554309/