How to find an element by href value using selenium python?

Question

I have href value of an anchor tag which only have href value as attribute. Now I want to find the element in the page which have same value as my href value and click it. I am unable to find any way of doing this using standard selenium methods.How can I do this? Basically these are the functions I found but it seems that I can't use any of these:

find_element_by_id
find_element_by_name
find_element_by_xpath
find_element_by_link_text
find_element_by_partial_link_text
find_element_by_tag_name
find_element_by_class_name
find_element_by_css_selector

ibodi · Accepted Answer · 2018-08-16 15:16:19Z

59

You can use find_element_by_xpath functionality.

driver.find_element_by_xpath('//a[@href="'+url+'"]')

edited Aug 16, 2018 at 15:16

ibodi

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answered Oct 15, 2015 at 18:06

Dmytro Pastovenskyi

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4 Comments

user3870509 Over a year ago

What if I have my url in a variable?

Dmytro Pastovenskyi Over a year ago

than just add it as a string there.

ccpizza Over a year ago

keep in mind that you might get an array returned if you have more than one link with the same URL

Jorma Rebane Over a year ago

This works very well, as long as the page has loaded all the contents

Duck Dodgers · Accepted Answer · 2019-01-22 10:22:09Z

15

You can try this:

driver.find_element_by_xpath('//a[contains(@href,"href")]')

edited Jan 22, 2019 at 10:22

Duck Dodgers

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answered Jan 22, 2019 at 10:03

Netra Shah

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Comments

rnevius · Accepted Answer · 2015-10-15 18:05:16Z

8

You would find the element by the CSS selector, as you would using vanilla CSS:

link = driver.find_element_by_css_selector('[href^=http://somelink.com/]')

You can also find the element by the link text:

link = driver.find_element_by_partial_link_text('somelink')

answered Oct 15, 2015 at 18:05

rnevius

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Comments

Tom Kuschel · Accepted Answer · 2022-02-13 17:30:27Z

4

When the button looks like

<a href="server_export.php" class="tab">&nbsp;Export</a>

and because possibly the text is translated, we cannot use the search for the text Export, so we have to search e.g. for server_export within the href and can e.g. click it with

driver.find_element(By.XPATH,'//a[contains(@href,"server_export")]').click()

Note: Using the recommended find_element(By.xxxxx, ...) format.

answered Feb 13, 2022 at 17:30

Tom Kuschel

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Comments

Daniel · Accepted Answer · 2020-10-23 17:24:50Z

1

I have a button called Your Profile with this html line

<a href="/registration/profile.html" rel="follow">Your Profile</a>

I have been trying everything so far find_element_by link, xpath, css selector with no luck. It returns error unable to find the element.

browser.find_element_by_partial_link_text("Your Profile")

browser.find_element_by_xpath('//a[contains(@href,"registration/profile.html")]')

Thanks

answered Oct 23, 2020 at 17:24

Daniel

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Comments

Adam Gonzales · Accepted Answer · 2023-04-11 02:24:22Z

0

In versions 4 and later of Selenium, use find_element() combined with a CSS selector and the By class:

from selenium import webdriver
from selenium.webdriver.common.by import By

jobs_link = driver.find_element(By.CSS_SELECTOR, "[href='https://www.linkedin.com/jobs/?']")
jobs_link.click()

answered Apr 11, 2023 at 2:24

Adam Gonzales

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