python selenium - click button and get page_source

Question

I currently click a button into a new window:

browser.execute_script("arguments[0].value = 'test';", browser.find_element_by_xpath('//*[@id="email-signup-email-address"]'))
input_button = browser.find_element_by_xpath('//*[@id="submittracking"]')
browser.execute_script('arguments[0].target="_blank";', input_button.find_element_by_xpath('./ancestor::form'))
browser.execute_script('arguments[0].click();', input_button)

but must time.sleep(10) before asking browser.page_source otherwise the page does not load fully

Other articles/posts suggest to use seleniums explicit wait but this asks to provide an element to wait for such as visibilityOfAllElements() or some specific element. I would like to get the ALL the page_source and am not looking to identify any elements.

How can I make browser.page_source work as it normally does (when not clicking into a new window)?

you can wait until state is completed, see this stackoverflow.com/questions/5868439/… — murali selenium
– murali selenium, Commented Dec 18, 2018 at 8:28
Do you mean, readyState? I may look into this further. Curious if it has a timeout built in. Also the link is mostly in Java — Rhys
– Rhys, Commented Dec 18, 2018 at 8:39

Andersson · Accepted Answer · 2018-12-18 09:02:04Z

1

Try to wait for page loaded (document.readyState == 'complete') as below:

from selenium.webdriver.support.ui import WebDriverWait

WebDriverWait(browser, 20).until(lambda browser: browser.execute_script("return document.readyState;") == "complete")

answered Dec 18, 2018 at 9:02

Andersson

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undetected Selenium · Accepted Answer · 2018-12-18 09:06:13Z

0

You can use some timeout for the page load like below:

driver.set_page_load_timeout(30)

edited Dec 18, 2018 at 9:06

undetected Selenium

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answered Dec 18, 2018 at 8:33

Kaushik

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