Python : is this local variable static?

Question

My First Attempt :

def generate_id():

    """ Create unique id of alphanumeric characters """
    i = 0
    id = ''
    while i!=10:
        id = id + random.choice(string.ascii_letters + string.digits)
        i+=1

    if check_unique(id):
           return id 

    id = generate_id()
    return id


def check_unique(id):
    """Check if id is unique"""
    try:
        instances = SomeModel.objects.get(id=id)
    except ObjectDoesNotExist:
        return True

    return False

Second Way :

def generate_id():

    """ Create unique id of alphanumeric characters """
    i = 0
    id = ''
    while i!=10:
        id = id + random.choice(string.ascii_letters + string.digits)
        i+=1

    if check_unique(id):
           return id 

    generate_id()



def check_unique(id):
    """Check if id is unique"""
    try:
        instances = SomeModel.objects.get(id=id)
    except ObjectDoesNotExist:
        return True

    return False

If I do it the second way , won't my logic of generating unique id's be wrong ? Because I might loose the id from the last call .

I am new to python and I don't know but I think my recursion concept looks messed up

Answer 1

Follow your code:

if check_unique(id):  # If this is `false`, you keep going
    return id 

generate_id()  # Now what? You call the function. Nothing gets returned.

If you want to create a unique ID, don't use recursion. Just use a while loop and generate new IDs as long as they're not unique:

characters = string.ascii_letters + string.digits

def generate_id(length=10):
    return ''.join(random.choice(characters) for i in range(length))

def generate_unique_id(length=10):
    id = generate_id(length)

    while not check_unique(id):
        id = generate_id(length)

    return id

Answer 2

In the second way you should return a the end of the generate_id function:

return generate_id()

I would also suggest to make an iteration instead of a recursive call... it's seems cleaner in this situation.