[FIXED] How to make the user register using unique code

Issue

I'm building a website and I'd like to limit how many users can register on it.

As you can see from the photo, I have a db in phpMyAdmin called website with two tables in it. One for the codes that I create and one for store the registered users. I'd have a normal register html form with a field for 'CODE'. I 'd like to stop the user from registering if the code doesn't exist or doesn't match the codes I have in the table. After they register I'd like the code to be deleted from the table but I can't figgure out how to do it. This is my code in HTML

<form class="login-form" method="POST" action="#">
        <input type="text" name="username" required="required">
        <input type="password" name ="password" required="required">
        <input type="text" name="code" required="required">
        <input type="submit" value="Register">
</form>

PHP

<?php
$host = "192.168.11.32:3306";
$user = "cyka";
$password = "blyat";
$db = "website";
$conn = mysqli_connect($host, $user, $password, $db);

if(isset($_POST['code'])){
    $userCode = $_POST['code'];
    $userUsername = $_POST['username'];
    $userPassword = $_POST['password'];
    $sql = "SELECT * FROM codes WHERE codice = '$userCode'";
    $result = mysqli_query($conn, $sql);

    if(mysqli_num_rows($result)==1){
        $sql = "INSERT INTO users (username, password) VALUES ('$username', '$password')";
        if ($conn->query($sql) === TRUE) {
            echo "Thank you for registering!";
        } else {
            echo "Errore: " . $sql . "<br>" . $conn->error;
            }
        header('location: main.php');
    } else {
        echo ("The code is not valid!");
        header('location: error.php');
    }
} 
?>

Solution

Submit the form via an ajax request upon hitting 'register' on your form.

Example of an ajax request to a back-end file:

$('#register-form').submit( function(e){
        e.preventDefault();
        $.ajax({
            url : '/includes/register.php',
            type : "POST",
            data : $('#register-form').serialize(),
            success : function(data) {
                if (data == 'true') {
                    //code if successful
                } else {
                    // code if not successful
                    }
                }
            }
        });

    });

All the values in the form will be posted to back-end file.

You can access the 'code' value that they have entered like this:

$theirCode = $_POST['code'];

Get an array of all your codes in your database - You can write a simple select all function:

function getCodes() {
    global $db;
        try {
        $stmt = $db->prepare("SELECT * FROM codes");
        $stmt->execute();
        return $stmt->fetchall();
    } catch (Exception $e) {
        echo $e->getMessage();
    }
}

$myCodes = getcodes();

All that you need to do now is just loop through your array of codes and check if their code matches with a code in your array.

$match = false;
foreach($myCodes as $code){
    if($theirCode == $code){
        $match = true;
        break;
    } 
}

if($match == true){
    echo 'true';
} else {
    echo 'false';
}

Your ajax request will be listening for true/false. Handel the outcome as you wish.

I do my explanation makes sense and is of help. Thanks

Answered By - Juan J