我的标题几乎全部说明了,我想制作一个登录表单,如果登录失败则不会刷新页面,但如果登录成功,它应该指引我到正确的页面:)
<?php
$errormessage = "";
session_start();
if (@$_SESSION['logged_in'] == TRUE) {
header("Location: index-users.php");
exit();
}
if ( isset($_POST['username']) && isset($_POST['password']) ) {
$username = $_POST['username'];
$password = $_POST['password'];
$db = new mysqli('localhost', 'root', '', 'dbname');
$query = "SELECT * FROM dbname_users WHERE username = '$username' AND password = '$password'";
$result = $db->query($query);
$rows = $result->num_rows;
if($rows == 1){
$_SESSION['logged_in'] = TRUE;
while ($row = $result->fetch_assoc()) {
$_SESSION['user_id'] = $row['ID'];
}
header("Location: index-users.php");
}
else{
$errormessage = "Forkert brugernavn eller adgangskode!";
}
}
?>
<form action="login.php" method="post">
<input type="text" name="username" placeholder="Username">
<input type="password" name="password" placeholder="Password">
<input type="submit" value="Login">
<p><?=$errormessage;?></p>
</form>
任何人都知道如何做到这一点? :)
答案 0 :(得分:3)
您需要将其与AJAX或jQuery代码结合使用。 它不仅仅是php!
检查一下: - jquery login form in div without refreshing whole page
答案 1 :(得分:-1)
function myBad() {
$.post("http://localhost/vivsang/logbase.php", {
login : $("#mylogin").serialize()
}, function (login) {
if (login == 'true') {
window.location = 'home.php';
} else {
alert('Wrong username or password!');
}
});
使用此jQuery代码进行登录。您必须先了解放置它的位置以及如何使用它。