import itertools

from explicit import waiter, XPATH
from selenium import webdriver
from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.support import expected_conditions as EC
from selenium.webdriver.common.by import By
from time import sleep

def login(driver):
    username = ""  # 
    password = ""  # 

    # Load page
    driver.get("https://www.instagram.com/accounts/login/")
    sleep(3)
    # Login
    driver.find_element_by_name("username").send_keys(username)
    driver.find_element_by_name("password").send_keys(password)
    submit = driver.find_element_by_tag_name('form')
    submit.submit()

    # Wait for the user dashboard page to load
    WebDriverWait(driver, 15).until(
        EC.presence_of_element_located((By.LINK_TEXT, "See All")))


def scrape_followers(driver, account):
    # Load account page
    driver.get("https://www.instagram.com/{0}/".format(account))

    # Click the 'Follower(s)' link
    # driver.find_element_by_partial_link_text("follower").click
    sleep(2)
    driver.find_element_by_partial_link_text("follower").click()

    # Wait for the followers modal to load
    waiter.find_element(driver, "//div[@role='dialog']", by=XPATH)
    allfoll = int(driver.find_element_by_xpath("//li[2]/a/span").text)
    # At this point a Followers modal pops open. If you immediately scroll to the bottom,
    # you hit a stopping point and a "See All Suggestions" link. If you fiddle with the
    # model by scrolling up and down, you can force it to load additional followers for
    # that person.

    # Now the modal will begin loading followers every time you scroll to the bottom.
    # Keep scrolling in a loop until you've hit the desired number of followers.
    # In this instance, I'm using a generator to return followers one-by-one
    follower_css = "ul div li:nth-child({}) a.notranslate"  # Taking advange of CSS's nth-child functionality
    for group in itertools.count(start=1, step=12):
        for follower_index in range(group, group + 12):
            if follower_index > allfoll:
                raise StopIteration
            yield waiter.find_element(driver, follower_css.format(follower_index)).text

        # Instagram loads followers 12 at a time. Find the last follower element
        # and scroll it into view, forcing instagram to load another 12
        # Even though we just found this elem in the previous for loop, there can
        # potentially be large amount of time between that call and this one,
        # and the element might have gone stale. Lets just re-acquire it to avoid
        # tha
        last_follower = waiter.find_element(driver, follower_css.format(group+11))
        driver.execute_script("arguments[0].scrollIntoView();", last_follower)


if __name__ == "__main__":
    account = ""  # 
    driver = webdriver.Firefox(executable_path="./geckodriver")
    try:
        login(driver)
        print('Followers of the "{}" account'.format(account))
        for count, follower in enumerate(scrape_followers(driver, account=account), 1):
            print("\t{:>3}: {}".format(count, follower))
    finally:
        driver.quit()