Wait until page is loaded with Selenium WebDriver

Question

I want to scrape all the data of a page implemented by a infinite scroll. The following python code works.

for i in range(100):
    driver.execute_script("window.scrollTo(0, document.body.scrollHeight);")
    time.sleep(5)

This means every time I scroll down to the bottom, I need to wait 5 seconds, which is generally enough for the page to finish loading the newly generated contents. But, this may not be time efficient. The page may finish loading the new contents within 5 seconds. How can I detect whether the page finished loading the new contents every time I scroll down? If I can detect this, I can scroll down again to see more contents once I know the page finished loading. This is more time efficient.

Answer 1

The webdriver will wait for a page to load by default via .get() method.

As you may be looking for some specific element as @user227215 said, you should use WebDriverWait to wait for an element located in your page:

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 selenium.common.exceptions import TimeoutException

browser = webdriver.Firefox()
browser.get("url")
delay = 3 # seconds
try:
    myElem = WebDriverWait(browser, delay).until(EC.presence_of_element_located((By.ID, 'IdOfMyElement')))
    print "Page is ready!"
except TimeoutException:
    print "Loading took too much time!"

I have used it for checking alerts. You can use any other type methods to find the locator.

EDIT 1:

I should mention that the webdriver will wait for a page to load by default. It does not wait for loading inside frames or for ajax requests. It means when you use .get('url'), your browser will wait until the page is completely loaded and then go to the next command in the code. But when you are posting an ajax request, webdriver does not wait and it's your responsibility to wait an appropriate amount of time for the page or a part of page to load; so there is a module named expected_conditions.

Answer 2

Here I did it using a rather simple form:

from selenium import webdriver
browser = webdriver.Firefox()
browser.get("url")
searchTxt=''
while not searchTxt:
    try:    
      searchTxt=browser.find_element_by_name('NAME OF ELEMENT')
      searchTxt.send_keys("USERNAME")
    except:continue

Answer 3

How about putting WebDriverWait in While loop and catching the exceptions.

from selenium import webdriver
from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.support import expected_conditions as EC
from selenium.common.exceptions import TimeoutException

browser = webdriver.Firefox()
browser.get("url")
delay = 3 # seconds
while True:
    try:
        WebDriverWait(browser, delay).until(EC.presence_of_element_located(browser.find_element_by_id('IdOfMyElement')))
        print "Page is ready!"
        break # it will break from the loop once the specific element will be present. 
    except TimeoutException:
        print "Loading took too much time!-Try again"

Answer 4

Find below 3 methods:

readyState

Checking page readyState (not reliable):

def page_has_loaded(self):
    self.log.info("Checking if {} page is loaded.".format(self.driver.current_url))
    page_state = self.driver.execute_script('return document.readyState;')
    return page_state == 'complete'

The wait_for helper function is good, but unfortunately click_through_to_new_page is open to the race condition where we manage to execute the script in the old page, before the browser has started processing the click, and page_has_loaded just returns true straight away.

id

Comparing new page ids with the old one:

def page_has_loaded_id(self):
    self.log.info("Checking if {} page is loaded.".format(self.driver.current_url))
    try:
        new_page = browser.find_element_by_tag_name('html')
        return new_page.id != old_page.id
    except NoSuchElementException:
        return False

It's possible that comparing ids is not as effective as waiting for stale reference exceptions.

staleness_of

Using staleness_of method:

@contextlib.contextmanager
def wait_for_page_load(self, timeout=10):
    self.log.debug("Waiting for page to load at {}.".format(self.driver.current_url))
    old_page = self.find_element_by_tag_name('html')
    yield
    WebDriverWait(self, timeout).until(staleness_of(old_page))

---

For more details, check Harry's blog.

Answer 5

From selenium/webdriver/support/wait.py

driver = ...
from selenium.webdriver.support.wait import WebDriverWait
element = WebDriverWait(driver, 10).until(
    lambda x: x.find_element_by_id("someId"))

Answer 6

On a side note, instead of scrolling down 100 times, you can check if there are no more modifications to the DOM (we are in the case of the bottom of the page being AJAX lazy-loaded)

def scrollDown(driver, value):
    driver.execute_script("window.scrollBy(0,"+str(value)+")")

# Scroll down the page
def scrollDownAllTheWay(driver):
    old_page = driver.page_source
    while True:
        logging.debug("Scrolling loop")
        for i in range(2):
            scrollDown(driver, 500)
            time.sleep(2)
        new_page = driver.page_source
        if new_page != old_page:
            old_page = new_page
        else:
            break
    return True