Find button class with Selenium on Python

Question

I scrapping the marketplace from this URL https://www.tokopedia.com/sunxin

I want to get the data from the button called Info Toko, if user click this will shown a pop up content:

The original source code for this button is:

<button class="css-rhf1fq-unf-btn e1ggruw00"><span>Info Toko</span></button>

I've tried to get the element by xpath, classname, link text, but still not working.

driver.find_element_by_link_text('Info Toko')

Always get error message like this

Any ideas how to get this element?

Mohamed Sulaimaan Sheriff · Accepted Answer · 2020-07-22 05:53:22Z

2

Try by using xpath selector

In which you want to change your selector code as shown below

driver.find_element_by_xpath('//div[@class="css-ais6tt"]//button[3]')

This will work

answered Jul 22, 2020 at 5:53

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PDHide · Accepted Answer · 2020-07-22 07:39:26Z

When ever you use a method read its documentation :

continue_link = driver.find_element_by_link_text('Continue')
continue_link = driver.find_element_by_partial_link_text('Conti')

This two element locators identifies the element only using link text.

In selenium a link is "an anchor tag" , an anchor tag is used wrap a href link

in your case its a button tag and not a anchor 'a' tag, so link text method won't work

use xpath , css or class

xpath:

driver.find_element_by_xpath('//button[text()="Info Toko"]')

css:

css to find using class

driver.find_element_by_css_selector('button.css-rhf1fq-unf-btn.e1ggruw00')

class:

driver.find_element_by_class_name('css-rhf1fq-unf-btn e1ggruw00')

2 Answers 2