using an xpath in selenium python

Question

I used firebug in Firefox to get a xpath for a link that did not have an ID assigned to it. The link is a javascript link with an image as the actual button. I'd like to be able to click this link but it's not working.

The actual xpath is '/html/body/div[2]/div/div/div[3]/div/div/table/tbody/tr[1]/td[2]/form/table/tbody/tr[1]/td/div[1]/div/table/thead/tr[2]/th[1]/a/img'

from selenium import webdriver
from selenium.webdriver.common.keys import Keys
driver = webdriver.Firefox()

URL = 'http://example.com'

driver.set_window_size(1024, 768)
driver.get(url)

link = driver.find_element_by_xpath(//*[@id="//html/body/div[2]/div/div/div[3]/div/div/‌table/tbody/tr[1]/td[2]/form/table/tbody/tr[1]/td/div[1]/div/table/thead/tr[2]/th‌[1]/a/img"])
for link in links:
 link.click()

I get the following error: is either invalid or does not result in a WebElement. The following error occurred:\nInvalidSelectorError: Unable to locate an element with the xpath expression //*[@id=/html/body/div[2]/div/div/div[3]/div/div/table/tbody/tr[1]/td[2]/form/table/tbody/tr[1]/td/div[1]/div/table/thead/tr[2]/th[1]/a/img because of the following error:\n[Exception... "The expression is not a legal expression." code: "12" nsresult: "0x805b0033 (SyntaxError)" location: "<unknown>"]' ; Stacktrace:

<a onclick="if(typeof functionx == 'function'){functionx(document.getElement…,'liststuff','');}return false" href="#">

   <img style="border: 0px;" onmouseover="this.src='/add-hover.gif';" onmouseout="this.src='/add.gif';" alt="Add" src="/icon-add.gif"></img>

</a>

Could you show the relevant html code that contains the link, or provide a link to the web-site? Thanks. — alecxe
– alecxe, Commented Jul 23, 2014 at 20:56
@alecxe sorry I can't provide the url because it's on an internal server to a company. — user2242044
– user2242044, Commented Jul 23, 2014 at 20:59
A guess: the <a> tag may have an img attribute instead of a nested <img> tag. If that's true, try changing a/img to a/@img at the end of your XPath. — unutbu
– unutbu, Commented Jul 23, 2014 at 21:03
@user2242044 the xpath you have is very fragile. Provide an html code so that we can see the link there. — alecxe
– alecxe, Commented Jul 23, 2014 at 21:04
@alecxe I added the element in the original post, but I can't add in the full html code — user2242044
– user2242044, Commented Jul 23, 2014 at 21:11

alecxe · Accepted Answer · 2014-07-23 21:24:10Z

4

Instead of having an absolute path to the element, rely on it's parent and attributes:

link = driver.find_element_by_xpath('//a[@onclick and @href = "#" and img/@alt = "Add"]')

edited Jul 23, 2014 at 21:24

answered Jul 23, 2014 at 21:12

alecxe

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4 Comments

user2242044 Over a year ago

this executes will now code issues, but does not open the page that gets opened when I click the link manually.

alecxe Over a year ago

@user2242044 I've updated the xpath expression, please check.

user2242044 Over a year ago

still same result. No errors but not opened link. PyCharm gives the following result Process finished with exit code 0

user2242044 Over a year ago

This worked! I accidentally removed the link.click() line. Thank you so much. Marking as best answer.

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