Clicking on svg using selenium python

Question

svg icon is clickable.

<div class="some-class">
    <svg aria-label="Search" class="some-icon" width="24" height="24" fill="#000" viewBox="0 0 24 24">
        <path d="M9.5,...,5 9.5,5Z">
        </path>
    </svg>
</div>

Sample code:

from selenium import webdriver

driver = webdriver.Chrome(CHROME_DRIVER_LOCATION)
driver.find_element_by_xpath('//*[@id="SearchForm"]/div[1]/span/div[1]/div[2]/svg/path').click()

Error:

no such element: Unable to locate element: {"method":"xpath","selector":"//*[@id="SearchForm"]/div[1]/span/div[1]/div[2]/svg/path"}

undetected Selenium · Accepted Answer · 2019-07-17 12:44:06Z

8

To click() on the svg icon you can use the following solution:

driver.find_element_by_xpath('//div[@class="some-class"]/*[name()="svg"][@aria-label="Search"]').click()

You can find a couple of relevant discussions in:

answered Jul 17, 2019 at 12:44

undetected Selenium

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Comments

Naresh Bogam · Accepted Answer · 2019-07-17 12:41:44Z

4

The "svg" elements are not from the XHTML namespace but belongs to SVG namespace. Hence you have to specify name()="svg" while constructing the xpath for svg tags. for example : "/*[name()='svg']/*[name()='path'] "

For your reference , please find below discussion How to click on SVG elements using XPath and Selenium WebDriver through Java

answered Jul 17, 2019 at 12:41

Naresh Bogam

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Comments

dbc · Accepted Answer · 2021-11-15 23:07:15Z

0

For my case the following worked:

driver.find_element_by_xpath('//div[@class="some-class"]/*[name()="svg"][@aria-label="Search"]').click()

edited Nov 15, 2021 at 23:07

dbc

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answered Nov 15, 2021 at 22:56

João Maniaudet

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