Selenium open pop up window [Python]

Question

I am trying to click a link by:

driver.find_element_by_css_selector("a[href='javascript:openhistory('AXS0077')']").click()

This works nice if the link opens in a new window but in this case the link actually opens a pop up window. When I try clicking the link with this method, using selenium it gives me an error:

Message: u"The given selector a[href='javascript:openhistory('AXS0077')'] is either invalid or does not result in a WebElement. The following error occurred:\nInvalidSelectorError: An invalid or illegal selector was specified"

Is this not the right way ? because I think there may be some different way to deal with pop windows.

Try driver.find_element_by_css_selector("""a[href='javascript:openhistory("AXS0077")']""").click() or driver.find_element_by_css_selector("""a[href="javascript:openhistory('AXS0077')"]""").click() — falsetru
– falsetru, Commented Jun 7, 2014 at 9:57

yurisich · Accepted Answer · 2014-06-07 13:25:47Z

2

+50

Your css selector could be more generic, perhaps:

driver.find_element_by_css_selector("a[href^='javascript']").click()

You've got all kinds of crazy overlapping quotation marks there. You're probably confusing it.

answered Jun 7, 2014 at 13:25

yurisich

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Padraic Cunningham · Accepted Answer · 2014-06-08 09:55:11Z

0

I have more success using find_by_xpath

Take this site as an example popups

I use firebug to inspect the element and get the xpath.

Then using the following works perfectly.

from selenium import webdriver

baseurl="http://www.globalrph.com/davescripts/popup.htm"

dr = webdriver.Firefox()
dr.get(baseurl)
dr.find_element_by_xpath("/html/body/div/center/table/tbody/tr[7]/td/div/table/tbody/tr/td[2]/div[1]/form/table/tbody/tr[4]/td[1]/a").click()

answered Jun 8, 2014 at 9:55

Padraic Cunningham

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