How to iterate a unordered list in Python Selenium

Question

I would like to iterate through the below elements in Python using Selenium webdriver.

<ul class="skills-section">
  <li class="endorse-item has-endorsements " data-endorsed-item-name="Ear Surgery">
  <li class="endorse-item has-endorsements " data-endorsed-item-name="Healthcare">
  <li class="endorse-item has-endorsements " data-endorsed-item-name="Hospitals">
  <li class="endorse-item has-endorsements " data-endorsed-item-name="Surgery">
  <li class="endorse-item has-endorsements " data-endorsed-item-name="Medical Education">
  <li class="endorse-item has-endorsements " data-endorsed-item-name="Pediatrics">
  <li class="endorse-item has-endorsements " data-endorsed-item-name="Treatment">
  <li class="endorse-item has-endorsements " data-endorsed-item-name="Public Health">
  <li class="endorse-item has-endorsements " data-endorsed-item-name="Patient Safety">
  <li class="endorse-item has-endorsements " data-endorsed-item-name="Emergency Medicine">
</ul>

What I tried?

skillsSection = a.find_element_by_xpath("//ul[contains(@class, 'skills-section')]")
skillsList = skillsSection.find_elements_by_tag_name("li")
for skill in skillsList:
    print skill.find_element_by_xpath("//span[contains(@class,endorse-item-name')]/a").text

But the problem is, it is always printing the first element value. The element is not incrementing.

Community · Accepted Answer · 2017-05-23 12:07:46Z

1

What you'll want to do is find the tag 'ul' with class='skills-section' and then iterate over the children.

skillsSection = a.find_element_by_xpath("//ul[contains(@class, 'skills-section')]")
for child in skillsSection.find_elements_by_xpath(".//*"):
    ...

See this answer too for more details: Selenium Python get all children elements

edited May 23, 2017 at 12:07

CommunityBot

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answered Jul 13, 2015 at 18:23

matthewatabet

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Comments

Evan Larson · Accepted Answer · 2016-03-20 19:10:05Z

0

I've been dealing with a similar issue recently where I could only print the first item in a list. I discovered a solution that requires multiple lines, and avoids using xpaths. For your code, you may want to replace:

print skill.find_element_by_xpath("//span[contains(@class,'endorse-item-name')]/a").text

with something like:

a1 = skill.find_element_by_class_name('endorse-item-name')
a2 = a1.find_element_by_tag_name('a')
print a2.text

Not as elegant as one line with xpath, but it (should) work.

answered Mar 20, 2016 at 19:10

Evan Larson

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