CODEWITHSUNDEEP
javascriptpythonseleniumselenium-webdriver
PatrickR400
PatrickR400

Reputation: 43

How do I get the onclick "action" of a link using Python and Selenium?

I am scraping a job board (eluta.ca) using Python and Selenium.

I'm able to return everything I want except for the link to the extended job description. In the bit of HTML below the job "Claims Adjuster"; links to "http://www.eluta.ca./direct/p?i=f1a7daa360e9468d5837d821c9d328ec"

<h2 class="title">

    <span class="lk-job-title" title="Claims Adjuster" onclick="enav2('./direct/p?i=f1a7daa360e9468d5837d821c9d328ec')"></span>

</h2>

I can find the element using the code

webdriver.find_element_by_css_selector("#organic-jobs .organic-job:nth-child("+str(Line)+") .title .lk-job-title")

retrieve the job via .text, navigate to the extended description via .click()

Am I missing sometingh obvious that would return all or part of "enav2('./direct/p?i=f1a7daa360e9468d5837d821c9d328ec')"

Thanks in advance

Upvotes: 1

Views: 2417

Answers (1)

alecxe
alecxe

Reputation: 473893

Use .get_attribute():

element = webdriver.find_element_by_css_selector("#organic-jobs .organic-job:nth-child("+str(Line)+") .title .lk-job-title")
print element.get_attribute('onclick')

Upvotes: 2

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