单击带有 Selenium 和 python 的 CSS 元素

Question

I am trying to click a button using Selenium.

Below is the code

<button class="_11f5fc88e3dec7bfec55f7f49d581d78-scss _238c8d0c29802f43d5fb66614d042cfa-scss" title="Play" aria-label="Play" data-testid="play-button" style xpath="1">

I tried to do this by the css selector:

searchButton = browser.find_elements_by_css_selector('[class="_11f5fc88e3dec7bfec55f7f49d581d78-scss _238c8d0c29802f43d5fb66614d042cfa-scss"]').click()

MY ERROR: AttributeError: 'list' object has no attribute 'click'

Am I approaching this wrong? Im not understanding the error.

Answer 1

Refer this link of how to use different selectors..
https://selenium-python.readthedocs.io/locating-elements.html

It is recommended to Copy css selector path by inspecting particular element on website then mouse right click -> copy -> copy css path

or by analyzing the code manually enter path

Here p is paragraph tag and class name is content

element = driver.find_element_by_css_selector('p.content')

Answer 2

AttributeError: 'list' object has no attribute 'click'

This error appear because you use .find_elements , it will return a list.

To solve this issue, try this approach:

1. Using .find_element , without s .
2. Or still using .find_elements but with index, like this:

browser.find_elements_by_css_selector('[class="_11f5fc88e3dec7bfec55f7f49d581d78-scss _238c8d0c29802f43d5fb66614d042cfa-scss"]')[0].click()