PHP Sign In - mysql_fetch_array returns NULL

Question

I am attempting to get a simple account sign-in setup and functioning; However, when running my 'signIn' function I seem to keep hitting 'Unsuccessful login'.

I used the var_dump function trying to locate the problem, and saw that var_dump($row['username']); die; was returning NULL. I feel that the problem might be originating from my mysql_query or mysql_fetch_array.

Any help is greatly appreciated.

<?php
    $host = 'localhost';
    $database = 'database_name';
    $username = 'username';
    $password = 'password';

    // Create connection
    $conn = mysql_connect($host, $username, $password);

    // Check connection
    if ($conn->connect_error) {
        die('Connection failed: ' . $conn->connect_error);
    }

    echo 'Connected successfully. ';

    mysql_select_db($database) or die('Unable to select database. ');

    $signIn = $_POST['signIn'];

    function signIn() {
        // Start session
        session_start();

        // Variables
        $ID = $_POST['username'];
        $Password = $_POST['password'];

        if (!empty($ID)) {
            $query = mysql_query('SELECT * FROM `signin` WHERE username = "$ID" AND password = "$Password"');
            $row = mysql_fetch_array($query, MYSQL_ASSOC);

            // Troubleshooting
            var_dump($row['username']); die;

            if (!empty($row['username']) AND !empty($row['password'])) {
                $_SESSION['username'] = $row['password'];

                echo "Successful login. ";
            } else {
                echo "Unsuccessful login. ";
            }
        }
    }

    if(isset($signIn)) {
        signIn();
    }
?>

Below is a screenshot of the current database table structure.

Answer 1

The problem is variable interpolation:

mysql_query('SELECT * FROM `signin` WHERE username = "$ID" AND password = "$Password"');

single quotes will not expand $ID and $Password to their values in the query, you need to use double quotes:

mysql_query("SELECT * FROM `signin` WHERE username = '$ID' AND password = '$Password'");