Python and Selenium - Get element that is visible

Question

Using Python 3.

Situation:

I select some elements with a certain xpath query.

There are many matches for that xpath query.

I want to grab the exact match corresponding with the element that is visible at the moment.

Notes:

There are always N matches (being N greater than 1.)

There is always only one match that is visible.

Actually, this is about pop-ups being displayed or not with javascript at certain moments.

Question:

How can iterate on all those results and know which one is visible by the user?

UPDATE:

The website is: go to website

If you wait a few seconds, there is a popup being displayed.

My xpath query is:

//div[@class='wrapper-code-reveal']//input[@class='code']

But there are 23 matches in this case.

How may I get the exact match that is being displayed?

I tried clicking it, which would give an exception when not visible.

codigos_descuento = driver.find_elements_by_xpath("//div[@class='wrapper-code-reveal']//input[@class='code']")

for codigo in codigos_descuento:
    try:
        codigo.click()
        codigo_descuento_texto = codigo.get_attribute('value')
    except:
        print(traceback.format_exc())
        continue

Paco H. · Accepted Answer · 2017-07-25 05:59:36Z

1

Here a working example with Selenium. Instead of fiddling with finding the correct one by visibility, I get the offer ID from the URL and use that to find the correct element.

from selenium import webdriver

driver = webdriver.Chrome()


def get_offer_id_from_url(url):
    offer_id = url.split('#')[1]
    offer_id = offer_id.split('-')[1]
    return offer_id


def get_discount_code(url, offer_id):
    offer_div_id = 'd-%s' % offer_id
    driver.get(url)
    discount_elem = driver.find_element_by_xpath(
        "//div[@id='%s']//input[@class='code']" % offer_div_id
    )
    discount_code = discount_elem.get_attribute('value')
    return discount_code


url = 'https://www.savoo.es/c-Alimentacion-codigo-promocional.html#p-5204957'
offer_id = get_offer_id_from_url(url)
discount_code = get_discount_code(url, offer_id)
print(discount_code)

answered Jul 25, 2017 at 5:59

Paco H.

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1 Comment

Álvaro N. Franz Over a year ago

Muchas gracias tito Paco. :) Wish you a great day.

Angelo · Accepted Answer · 2018-08-21 13:50:49Z

0

My answer comes a little late but I think one of the ways to check if the element is visible is simply by checking its location on the screen. visible_elem = None

for elem in codigos_descuento:
    if elem.location["x"] > 0 and elem.location["y"] >0:
        visible_elem = elem
        print("Visible Element found!")
        break

answered Aug 21, 2018 at 13:50

Angelo

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