Selenium making a list with the get_attributes method

Question

I'm trying to get random wikipedia articles with the wikipedia API and then get all their href links which I did by:

elems = driver.find_elements_by_xpath("//a[@href]")

Then I made:

for elem in elems:
    elements = []

    elements.append(elem.get_attribute("href"))

In order to only get the href links from the received list with the previously executed command. I only got a single link which is:

['https://www.mediawiki.org/']

I tried to do:

for elem in elems:
    elements = []
    elements.extend(elem.get_attribute("href"))

Which resulted in this output:

['h', 't', 't', 'p', 's', ':', '/', '/', 'w', 'w', 'w', '.', 'm', 'e', 'd', 'i', 'a', 'w', 'i', 'k', 'i', '.', 'o', 'r', 'g', '/']

Clearly put: I want to take all href links from a random wikipedia article then put them in a single list called elements which I will later use for a for loop.

Thanks in advance.

Rest of my code if it is needed:

import wikipedia
import selenium
from selenium import webdriver

#This will get a random wikipedia article
random_article = wikipedia.random(1)
#Get the articles URL
url = wikipedia.page(random_article).url

driver = webdriver.Firefox()
driver.get(url)

elems = driver.find_elements_by_xpath("//a[@href]")
for elem in elems:
    elements = []
    elements.extend(elem.get_attribute("href"))

print(elements)

Thx for accepting my ans as the best ans! Can u also upvote my ans? Thanks! — Sushil
– Sushil, Commented Nov 7, 2020 at 7:43

Sushil · Accepted Answer · 2020-11-07 07:35:48Z

1

You have to initialise the elements list outside the for loop, so this is how for loop should look like:

elements = []

for elem in elems:
    elements.append(elem.get_attribute("href"))

Full code:

import wikipedia
import selenium
from selenium import webdriver

#This will get a random wikipedia article
random_article = wikipedia.random(1)
#Get the articles URL
url = wikipedia.page(random_article).url

driver = webdriver.Firefox()
driver.get(url)

elems = driver.find_elements_by_xpath("//a[@href]")
elements = []
for elem in elems:
    elements.append(elem.get_attribute("href"))

print(elements)

answered Nov 7, 2020 at 7:35

Sushil

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