Unable to open href link in selenium python - selenium

Question

I have a array called accounts which gets all the href's i want, i then want to open each of these, i have tried the following code
accounts = self.driver.find_elements_by_xpath("//a[contains(#href, '/signin?')]")
for account in accounts:
self.driver.get(account)
time.sleep(3)
But returns
selenium.common.exceptions.InvalidArgumentException: Message: invalid argument: 'url' must be a string
(Session info: chrome=80.0.3987.132)

Answer 1

You are fetching the list of web elements, so you need to first fetch the href attribute from those web elements and then hit them.
You can do it like:
accounts = self.driver.find_elements_by_xpath("//a[contains(#href, '/signin?')]")
for account in accounts:
self.driver.get(account.get_attribute("href"))
time.sleep(3)

Answer 2

To open the href, instead of the WebElement you need to invoke get() passing the href attribute inducing WebDriverWait for the visibility_of_all_elements_located() and you can use either of the following Locator Strategies:
Using CSS_SELECTOR:
accounts = [my_elem.get_attribute("href") for my_elem in WebDriverWait(driver, 5).until(EC.visibility_of_all_elements_located((By.CSS_SELECTOR, "a[href*='/signin?']")))]
for account in accounts:
self.driver.get(account)
Using XPATH:
accounts = [my_elem.get_attribute("href") for my_elem in WebDriverWait(driver, 5).until(EC.visibility_of_all_elements_located((By.XPATH, "//a[contains(#href, '/signin?')]")))]
for account in accounts:
self.driver.get(account)
Note : You have to add the following imports :
from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.common.by import By
from selenium.webdriver.support import expected_conditions as EC