CODEWITHSUNDEEP
pythonseleniumselenium-webdriver
Vipul
Vipul

Reputation: 4168

Selenium open pop up window [Python]

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.

Upvotes: 4

Views: 1819

Answers (2)

Padraic Cunningham
Padraic Cunningham

Reputation: 180411

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()

Upvotes: 0

yurisich
yurisich

Reputation: 7119

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.

Upvotes: 2

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