如果我在自身内部重新运行一个函数，PDO就无法运行

Question

I'm trying to create a system, where my website generates an unique code (current date + 5 random characters) and that code is transferred to a table in my database.

Before function generateNumber() can insert the unique code into the database, it has to check if the code already exist in the database.

If the code doesn't exist, my function works flawlessly. But the problem is when the code can already be found on the database, my website just doesn't do anything (it should just re-run the function).

function generateNumber()
{
    global $conn;

    $rand = strtoupper(substr(uniqid(sha1(time())),0,5));
    $result = date("Ydm") . $rand;

    $SQL = $conn->query("SELECT code FROM test WHERE code='$result'");
    $c = $SQL->fetch(PDO::FETCH_ASSOC);

    if ($c['code'] > 0) { // test if $result is already in the database
        generateNumber();
    } else {
        $sql2 = "INSERT INTO test (code) VALUES (?)";
        $stmt2 = $conn->prepare($sql2);
        $stmt2->execute([$result]);
        return $result;
    }
}

try {
    $conn = new PDO("sqlite:db"/*, $username, $password*/);
    $conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);

    echo generateNumber();
}

catch(PDOException $e) {
    echo "Error:" . $e->getMessage();
}

$conn = null;
?>

There are no error messages in the console, but I suspect the problem is this part of the code:

if ($c['code'] > 0) { // test if $result is already in the database
    generateNumber();
} 

Any idea how I can write this function in a better way?

Answer 1

Solution :

if ($c['code'] > 0) { // test if $result is already in the database
// if the code exists in db the function return nothing 
// because you are missing a return :
    return generateNumber();

}