XPath locate child following sibling

Question

I am using Selenium for Python 2.7.10.

With XPath, I would like to locate the link in a href, following the sibling to minimal-list__title (i.e. I'm looking for the child beneath minimal-list__value). Which XPath should I use?

<span class="minimal-list__title">ETF Home Page:</span>
<span class="minimal-list__value">
    <a href="http://www.robostoxetfs.com/">ROBO</a>

This is the current attempt:

from selenium import webdriver as driver
from selenium.common.exceptions import NoSuchElementException

def get_link(driver, key):
    key = key + ":"
    try:
        find_value = driver.find_element_by_xpath("//span[@class='minimal-list__title' and . = '%s']/following-sibling::span/*[1]::a" % key).text
    except NoSuchElementException:
        return None
    else:
        value = re.search(r"(.+)", find_value).group().encode("utf-8")
        return value

website = get_link(driver, "ETF Home Page")
print "Website: %s" % website

Note that I am specifically interested in a XPath that gets the link from the child of the following sibling. This is because the function above uses "ETF Home Page:" in the web code as an identifier for what to search for.

alecxe · Accepted Answer · 2015-08-05T15:24:44.913

2

You're almost correct:

//span[@class = "minimal-list__title" and . = "ETF Home Page:"]/following-sibling::span/a

Note that you don't need to worry about multiple elements matching the locator since you are using find_element_by_xpath() and it would give you the first matching element.

Though, if it would makes sense in your case and you know the "ROBO" label beforehand:

driver.find_element_by_link_text("ROBO")

To get an attribute value, use get_attribute():

find_value = driver.find_element_by_xpath('//span[@class = "minimal-list__title" and . = "ETF Home Page:"]/following-sibling::span/a').get_attribute("href")

edited Aug 05 '15 at 15:24

answered Aug 05 '15 at 14:55

alecxe

462,703
120
1,088
1,195

Thanks :) This gives me the element text `ROBO`, but I would like to get the url in `href`. Something like `/following-sibling::span/a[@href]`? Produces the same result however. – P A N Aug 05 '15 at 15:10

score 0 · Answer 2 · edited Feb 15 '17 at 05:58

0

String e = driver.findElement(By.xpath("//*[contains(@class,"minimal-list__value")]/a)).getAttribute("href");

//*[contains(@class,"minimal-list__value")]/a is the xpath, the getAttribute will give you the desired result.

edited Feb 15 '17 at 05:58

MarmiK

5,639
6
40
49

answered Feb 15 '17 at 04:02

akhilesh gulati

128
8

1

When giving an answer it is preferable to give [some explanation as to WHY your answer](http://stackoverflow.com/help/how-to-answer) is the one. – Stephen Rauch Feb 15 '17 at 04:33
//*[contains(@class,"minimal-list__value")]/a it is the xpath the get attribute will give you the desired result. – akhilesh gulati Feb 15 '17 at 05:21
The explanation should be in the post. There is an edit button above and to the left of this comment. – Stephen Rauch Feb 15 '17 at 05:22

score 0 · Answer 3 · answered Sep 01 '20 at 20:08

Based on the text ETF Home Page: to extract the link http://www.robostoxetfs.com/ from the child node of the following sibling you can use either of the following xpath based Locator Strategies:

Using xpath and following-sibling:

print(driver.find_element_by_xpath("//span[text()='ETF Home Page:']//following-sibling::span/a").get_attribute("href"))

Using xpath and following:

print(driver.find_element_by_xpath("//span[text()='ETF Home Page:']//following::span/a").get_attribute("href"))

XPath locate child following sibling

3 Answers3

Linked