How to open a new window on a browser using Selenium WebDriver for python?


I am attempting to open a new tab OR a new window in a browser using selenium for python. It is of little importance if a new tab or new window is opened, it is only important that a second instance of the browser is opened.

I have tried several different methods already and none have succeeded.

  1. Switching to a window that does not exist with hopes that it would then open a new window upon failure to locate said window:


  2. Iterating through open windows (although there is currently only one)

    for handle in driver.window_handles:
  3. Attempting to simulate a keyboard key press

    from selenium.webdriver.common.keys import Keys
    driver.send_keys(Keys.CONTROL + 'T')

The problem with this one in particular was that it does not seem possible to send keys directly to the browser, only to a specific element like this:

driver.find_element_by_id('elementID').send_keys(Keys.CONTROL + 'T')

However, when a command such as this is sent to an element, it appears to do absolutely nothing. I attempted to locate the topmost HTML element on the page and send the keys to that, but was again met with failure:

driver.find_element_by_id('wrapper').send_keys(Keys.CONTROL + 'T')

Another version of this I found online, and was not able to verify its validity or lack thereof because I’m not sure what class/module which needs importing

act = ActionChains(driver)

Something very similar with different syntax (I’m not sure if one or both of these is correct syntax)



How about you do something like this

driver = webdriver.Firefox() #First FF window
second_driver = webdriver.Firefox() #The new window you wanted to open

Depending on which window you want to interact with, you send commands accordingly

print driver.title #to interact with the first driver
print second_driver.title #to interact with the second driver

For all down voters:

The OP asked for “it is only important that a second instance of the browser is opened.“. This answer does not encompass ALL possible requirements of each and everyone’s use cases.
The other answers below may suit your particular need.

Answered By – Amey

This Answer collected from stackoverflow, is licensed under cc by-sa 2.5 , cc by-sa 3.0 and cc by-sa 4.0

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