I'm trying to change CSS style of an element (example: from "visibility: hidden;" to "visibility: visible;") using selenium .execute_script. (any other method through selenium+python would be accepted gracefully).
my code:
driver = webdriver.Firefox()
driver.get("http://www.example.com")
elem = driver.find_element_by_id('copy_link')
elem.execute_script( area of my problem )
what do i need to do in order to play with the CSS of the webpage ?
Here is an example without using any jQuery. It will hide Google's logo.
from selenium import webdriver
driver = webdriver.Firefox()
driver.get("http://www.google.com")
driver.execute_script("document.getElementById('lga').style.display = 'none';")
The same idea could be used to show a hidden element by setting .style.display to "block", for example.
<iframe>? Look at driver.page_source to see whether the element you wish to show or hide is actually present in the DOM. Also, ensuring that you have the latest version of selenium might be worthwhile.div so that we can actually run your code. There is nothing else I can say without seeing the problem myself.Here is solution i found using document style sheets. This way is great because you can also add pseudo class styling.
script = 'document.styleSheets[0].insertRule("button:focus {background-color: red !important;}", 0 )'
driver.execute_script(script)
String in execute_script() is JS code you want to run (docs).
If you use jQuery it can be just
driver.execute_script("$('#copy_link').css('visibility', 'visible');")
If you catch the element by its xpath, then you can change or add styles to it as
location_selector = driver.find_element(By.XPATH, '//select[@data-placeholder="Return Locations"]')
driver.execute_script("arguments[0].style.display = 'block';", location_selector)
