Selenium with Python: Stale Element Reference Exception

Add these imports:

from selenium.webdriver.common.by import By
from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.support import expected_conditions

Change these lines

inputbox.send_keys(Keys.ENTER)
self.check_for_row('1: Buy peacock feathers')

to:

inputbox.send_keys(Keys.ENTER)

WebDriverWait(self.browser, 10).until(
            expected_conditions.text_to_be_present_in_element(
                (By.ID, 'id_list_table'), 'Buy peacock feathers'))

self.check_for_row('1: Buy peacock feathers')

This replaces the time.sleep(1) with something more "reasonable"

I have been using selenium for a while now so I understand the struggles of the Stale Element Exception. While not perfect, selenium provides a series of "wait" commands to allow for the website to load complete. Unfortunately, its not perfect as loading can take different time on each run, but these are the tools provided by selenium.

I haven't worked in python but have worked on java/selenium. But,I can give you the idea to overcome staleness.

Generally we will be getting the Stale Exception if the element attributes or something is changed after initiating the webelement. For example, in some cases if user tries to click on the same element on the same page but after page refresh, gets staleelement exception.

To overcome this, we can create the fresh webelement in case if the page is changed or refreshed. Below code can give you some idea.(It's in java but the concept will be same)

Example:

webElement element = driver.findElement(by.xpath("//*[@id='StackOverflow']"));
element.click();
//page is refreshed
element.click();//This will obviously throw stale exception

To overcome this, we can store the xpath in some string and use it create a fresh webelement as we go.

String xpath = "//*[@id='StackOverflow']";
driver.findElement(by.xpath(xpath)).click();
//page has been refreshed. Now create a new element and work on it
driver.fineElement(by.xpath(xpath)).click();   //This works

Another example:

for(int i = 0; i<5; i++)
{
  String value = driver.findElement(by.xpath("//.....["+i+"]")).getText);
  System.out.println(value);
} 

Hope this helps you. Thanks

To prevent an element to become stale, place a new element on the current page, hit the link and wait until the element is not available anymore. Then wait for an element on the new page to appear

    script_to_execute = """
            var new_element = document.createElement('span');
            new_element.setAttribute('id', 'wait4me');
            document.body.appendChild(new_element);"""
    self.driver.execute_script(script_to_execute)
    self.driver.find_element_by_xpath("{}".format(locator)).click()
    WebDriverWait(self.driver, self.time_out).until (
                lambda x: not x.find_elements_by_id("wait4me")) 

This issue happens when the loop starts before an updated page has fully loaded. Especially when you update a page in an application or a form. One workaround is to place an element on the current page, then update and use the WebDriverWait statement until the element is not found anymore. Then start your loop. (Otherwise the reload happens during the loop...)

I read the same book as you do and encountered the same problem (solutions from this page didn't work for me). Here's how I resolved it.

Problem

Exception is thrown whenever you try to access a stale object. So we have to wait for situation when this exception is NOT thrown anymore.

My solution

I created method that waits for my actions until they pass

from selenium.common.exceptions import StaleElementReferenceException
[...]
def stale_aware_for_action(self, action):
    while(True):
        try:
            action()
            break
        except StaleElementReferenceException:
            continue

And in test method I defined actions that I want to wait to finish:

def test_can_start_a_list_and_retrieve_it_later(self):
    [...]
    def insert_second_item_to_inputbox():
        inputbox = self.browser.find_element_by_id('id_new_item')
        inputbox.send_keys('Use peacock feathers to make a fly')
        inputbox.send_keys(Keys.ENTER)

    self.stale_aware_for_action(insert_second_item_to_inputbox)

    def check_for_first_item():
        self.check_for_row_in_list_table('1: Buy peacock feathers')
    def check_for_second_item():
        self.check_for_row_in_list_table('2: Use peacock feathers to make a fly')

    self.stale_aware_for_action(check_for_first_item)
    self.stale_aware_for_action(check_for_second_item)