Exchange $_POST for isset

Question

Could you help me Exchange $_POST for isset my code ...

$login = $db->real_escape_string(htmlspecialchars(trim($_POST['login'])));
$password = $_POST['password'];

I put:

if(isset($_POST['login'])) $login = $db->real_escape_string(htmlspecialchars(trim($_POST['login']))); else $login = your default value;

and now i have error in next line:

error: Parse error: syntax error, unexpected 'if' (T_IF) in C:\xampp\htdocs\CDR\login.php on line 31

if ($_POST) {
$errors = array();

if (empty($login) || empty($password)) {
    $errors[] = '......';
}

$auth = $user->auth($login, $password);
if (!$auth) {
    $errors[] = '......';
}


if (empty($errors)) {

    $_SESSION['user_id'] = $auth;

    echo '<p class="success">.....<a href="index.php">....</a></p>';
} else {
    foreach ($errors as $error) {
        echo '<p class="error">'.$error.'</p>';
    }
}

}

What is the error / problem ? More details could be usefull. — ThinkTank, Mar 03 '16 at 09:41
`$_POST` and `isset` are not interchangeable. Be more specific. — apokryfos, Mar 03 '16 at 09:43
I have error Notice: Undefined variable: db in C:\xampp\htdocs\CDR\login.php on line 26 Fatal error: Call to a member function real_escape_string() on null in C:\xampp\htdocs\CDR\login.php on line 26 and i need change $_POST to isset — Hemikalny, Mar 03 '16 at 09:46
$login = $db->real_escape_string(htmlspecialchars(trim($_POST['login']))); $password = $_POST['password']; //czy tak jest poprawnie ?? if ( isset($_Post["login"]) ){ $login =mysql_real_escape_string(htmlspecialchars($_POST["password"])); } else { $password=''; } — Hemikalny, Mar 03 '16 at 09:50
For crying out loud, you're barking up the wrong tree entirely... http://stackoverflow.com/a/12769983/476, http://stackoverflow.com/a/12778634/476 — deceze, Mar 03 '16 at 09:55
@Hemikalny Please see [ask] it's unclear what your question is. Also make sure you put your information in the [edit] question and not the comments. — Rizier123, Mar 03 '16 at 09:59

score 0 · Answer 1 · answered Mar 03 '16 at 09:49

0

You probably want to check if your login post variable has any value.. You can also do this for your password post variable:

if(isset($_POST['login']))
 $login = $db->real_escape_string(htmlspecialchars(trim($_POST['login'])));
else
 $login = your default value;

answered Mar 03 '16 at 09:49

James Joyce Alano

703
3
13

Now i heve error in next line – Hemikalny Mar 03 '16 at 10:07
'if ($_POST) { $errors = array(); if (empty($login) || empty($password)) { $errors[] = 'Wypełnij wszystkie pola'; } $auth = $user->auth($login, $password); if (!$auth) { $errors[] = 'Użytkownik o podanym loginie i haśle nie istnieje'; } if (empty($errors)) { $_SESSION['user_id'] = $auth; echo '
Zostałeś zalogowany. Możesz przejść na stronę główną
'; } else { foreach ($errors as $error) { echo '
'.$error.'
'; } } } ' – Hemikalny Mar 03 '16 at 10:08

score 0 · Accepted Answer · answered Mar 03 '16 at 09:49

I think you mean checking the post variables:

$login = isset($_POST['login']) ? $_POST['login'] : "";
$password = isset($_POST['password']) ? $_POST['password'] : "";

Also, instead of doing escaping the way you use in your question, consider using PDO (http://php.net/manual/en/book.pdo.php).

Exchange $_POST for isset

2 Answers2