xRahim
Hectopat
- Katılım
- 22 Aralık 2019
- Mesajlar
- 200
Daha fazla
- Cinsiyet
- Erkek
Herkese merhaba. Aşağıdakı kodlamada hata nerede? PHP 8 kurulu sistemde.
PHP:
<?php
include 'db_conn.php';
?>
<!DOCTYPE html>
<html>
<head>
<meta charset=utf-8>
<link rel="stylesheet" type="text/css" href="css/style.css">
<title>Reg</title>
</head>
<body>
<?php
if(empty($_POST['submit']))
{
?>
<form id="reg_form" action="" method="POST">
Login
<input type="text" name="login" required><br>
Parol
<input type="password" name="pass" required><br>
<input type="submit" name="submit" value="Reg">
</form>
<?php
}
else
{
$err = array();
$post_login = trim(htmlspecialchars(($_POST['login'])));
$sql = "SELECT COUNT(*) FROM users WHERE login='".$post_login."'";
$result = $conn->query($sql);
$row = mysqli_fetch_assoc($result);
if($result != null) ///Bu kısımda hata var
{
$err[] = "Bu isimde kullanici kayitli";
}
if(strlen($_POST['login']) < 3 or strlen($_POST['login']) > 30)
{
$err[] = "Login 3-30 karakter olmali";
}
if(strlen($_POST['pass']) < 3 or strlen($_POST['pass']) > 30)
{
$err[] = "Parola 3-30 karakter olmali";
}
if(count($err) == 0)
{
$login = trim(htmlspecialchars(($_POST['login'])));
$pass = trim(htmlspecialchars(($_POST['pass'])));
$sql = "INSERT INTO users (login, pass)
VALUES ('".$login."', '".$pass."')";
echo'Kayit basarili <br>';
echo'
<u>Login:</u> '.$login.'<br />
<u>Parola:</u> '.$pass.'<br />';
echo'<a href="auth.php"><input value="Giris" /></a>';
}else{
foreach($err as $error)
{
$errors .= ++$n.') '.$error.'; <br>'; ///Burada 1,2 ve 3 hata olmasını istiyorum
}
?>
Hata:<br>
<?php echo $errors; ?><br>
<a href="javascript:history.back()" return true"><input value="Geri" type="button"></a>
<?php
}
}
?>
</body>
</html>