Why "Undefined variable" error?

Question

I am learning PHP by myself. I am using Slim framework with Medoo and creating a login page and when executing login page there is no error. but when calling "call_query" function error occur saying :

Undefined variable: database in C:\xampp\htdocs\class\blog\pages\login.php on line 8

and

Fatal error: Call to a member function select() on a non-object in C:\xampp\htdocs\class\blog\pages\login.php on line 8

In code i have created an object $database then using select method on it.

<?php
require_once 'medoo.php';
$database = new medoo();
session_start();
$login_error = $br = null;

function call_query(){
    $data = $database->select('user', 'user_name', [
        'user_name'=>$_POST['u_name'],
        'password'=>$_POST['pass']
        ]);
    if (count($data) == 1) {
        header("location : localhost/class/blog");
    }else{
        $login_error = "Wrong Username or Password";
        $br = "<br><br>";
    }
}

if ($_SERVER['REQUEST_METHOD'] == 'POST' && $_POST['submit'] == 'submit'){
    if (isset($_POST['u_name']) && isset($_POST['pass']) && !empty($_POST['u_name']) && !empty($_POST['pass'])) {
        if (isset($_SESSION['username']) && isset($_SESSION['password'])) {
            if ($_SESSION['username'] == $_POST['u_name'] && $_SESSION['password'] == $_POST['pass']) {
                header("location : localhost/class/blog");
            }else{
                call_query();
            }
        }else{
            call_query();
        }
    }else{
        $login_error = "Enter both Username and Password to login.";
        $br = "<br><br>";
    }
}

echo '<form action = "pages/login.php" method = "post">
        <label>Username : </label><input type ="text" name ="u_name"><br><br>
        <label>Password : </label><input type = "password" name = "pass"><br><br>
        '.$login_error.$br.'
        <input type ="submit" name = "submit" value = "submit">
     </form>';
?>

Pehaps maybe $database it's not available inside call_query() function. Check @jeroen link. — lmarcelocc, Feb 24 '15 at 10:13
line 8 is where ($data = $database->select('user', 'user_name', [) in call_query function() — Nishant Bhakta, Feb 24 '15 at 10:20

score 0 · Answer 1 · answered Feb 24 '15 at 12:07

Making $database as global variable worked. Thanks @lmarcelocc, @jeroen, @Md. Sahadat Hossain.

Now code is :

<?php
require_once 'medoo.php';
$database = new medoo();
session_start();
$login_error = $br = null;

function call_query($username, $password){
    global $database, $login_error, $br;
    $data = $database->select('user', 'user_name', ['user_name'=>$username, 'password' => $password]);
    if (!empty($data)) {
        print_r($data);
        $_SESSION['username'] = $username;
        $_SESSION['password'] = $password;
        header("location: localhost/class/blog");
    }
}

if ($_SERVER['REQUEST_METHOD'] == 'POST' && $_POST['submit'] == 'submit'){
    if (isset($_POST['u_name']) && isset($_POST['pass']) && !empty($_POST['u_name']) && !empty($_POST['pass'])) {
        if (isset($_SESSION['username']) && isset($_SESSION['password'])) {
            if ($_SESSION['username'] == $_POST['u_name'] && $_SESSION['password'] == $_POST['pass']) {
                $_SESSION['username'] = $_POST['u_name'];
                $_SESSION['password'] = $_POST['pass'];
                header("location: /");
                echo "logged by session";
            }else{
                call_query($_POST['u_name'], $_POST['pass']);
                $login_error = "Invalid Username or Password.";
                $br = "<br><br>";
            }
        }else{
            call_query($_POST['u_name'], $_POST['pass']);
            $login_error = "Invalied Username or Password.";
            $br = "<br><br>";
        }
    }else{
        $login_error = "Enter both Username and Password to login.";
        $br = "<br><br>";
    }
}

echo '<form action = "pages/login.php" method = "post">
        <label>Username : </label><input type ="text" name ="u_name"><br><br>
        <label>Password : </label><input type = "password" name = "pass"><br><br>
        '.$login_error.$br.'
        <input type ="submit" name = "submit" value = "submit">
     </form>';
?>

Why "Undefined variable" error?

1 Answers1