How to identify an element using webdriver in python for http link

Question

How to identify the link, I have inspected the elements which are as below :

<div class="vmKOT" role="navigation">
<a class="Ml68il" href="https://www.google.com" aria-label="Search" data-track-as="Welcome Header Search"></a>
<a class="WaidDw" href="https://mail.google.com" aria-label="Mail" data-track-as="Welcome Header Mail"></a>
<a class="a4KP9d" href="https://maps.google.com" aria-label="Maps" data-track-as="Welcome Header Maps"></a>
<a class="QJOPee" href="https://www.youtube.com" aria-label="YouTube" data-track-as="Welcome Header YouTube"></a>
</div>

I want to identify the class WaidDw or href and click it using python.

Your question seems incomplete. Did you forget to add something or accidentally removed something? — koceeng, Feb 17 '17 at 07:59

score 2 · Answer 1 · edited Feb 17 '17 at 08:34

2

You can try

driver.find_element_by_class_name('WaidDw').click()

or

driver.find_element_by_xpath('//a[@href="https://mail.google.com" and @aria-label="Mail"]').click()

edited Feb 17 '17 at 08:34

Naveen Kumar R B

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answered Feb 17 '17 at 08:32

Andersson

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score 0 · Answer 2 · edited May 23 '17 at 11:53

In your provided HTML all attribute's values are unique, you can locate easily that element by using their attribute value.

As your question points to locate this <a class="WaidDw" href="https://mail.google.com" aria-label="Mail" data-track-as="Welcome Header Mail"></a> element. I'm providing you multiple cssSelectors which can work easily to identify the same element as below :-

a.WaidDw
a.WaidDw[href='https://mail.google.com']
a.WaidDw[aria-label='Mail']
a.WaidDw[data-track-as='Welcome Header Mail']
a.WaidDw[href='https://mail.google.com'][aria-label='Mail']
a.WaidDw[href='https://mail.google.com'][aria-label='Mail'][data-track-as='Welcome Header Mail']

Note :- Keep in practice (priority) to use cssSelector instead xpath if possible, because cssSelectors perform far better than xpath

Locating Element by CSS Selectors using python :-

element = driver.find_element_by_css_selector('use any one of the given above css selector')

Clicks the element :-

element.click()

Reference link :-

How to identify an element using webdriver in python for http link

2 Answers2