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This is my code:

myproject.py

userName.send_keys(user_name)
passWord.send_keys(pass_word)

Button.click()

print("1")

When I do Button.click(), the pages can successfully JUMP but it CANNOT continue to the NEXT STEP, even it is just a print() statement.

The next page is displayed like the image below.

enter image description here enter image description here

In the next page, it has alert windows up there, and I cannot switch to it. The program gave me an error message:

[20120:10844:0529/150013.017:ERROR:textfield.cc(1773)] NOT IMPLEMENTED

And I have used submit() too, it also did not change. I am really confused about this issue so I checked the lib of selenium:

webelement.py from Selenium

def click(self):
    """Clicks the element."""
    print("1")
    self._execute(Command.CLICK_ELEMENT)
    print("2")

Somehow the print("2") cannot be printed.

If someone knows the reason for this, I will appreciate it if you can tell me.

Gino Mempin
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Was me It
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  • It is blocking at the alert (I have tried to switch to it without any luck), could you check this solution https://stackoverflow.com/questions/11522434/how-to-handle-login-pop-up-window-using-selenium-webdriver. – Nic Laforge May 30 '19 at 02:11
  • Suggestion in the link is to bypass the login page/ authentication pop-up. Use ```driver.get("https://username:password@yourUrl")```. yourUrl is not your login page but the next one, which is the one halting the click. – Nic Laforge May 30 '19 at 02:45
  • It not work, because the link is not changed whether login or not, – Was me It May 30 '19 at 02:49
  • Got it, this is I know issue from chromedriver: https://bugs.chromium.org/p/chromedriver/issues/detail?id=1917&q=authentication&colspec=ID%20Status%20Pri%20Owner%20Summary. Can you run with FireFox? I am able to switch to the authentication pop-up. – Nic Laforge May 30 '19 at 02:56
  • Thanks a lot! I never thought it will be a bug – Was me It May 30 '19 at 05:46

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