[python][selenium] on-screen position of element

问题:

Hello i would like to know the on-screen position of some element. I know how to get the position of element in python selenium webriver but how to get offset from left-top corner of screen?

回答1:

I guess it's not possible to define distance from top-left corner of browser window to top-level corner of screen with just selenium. But you can try to implement following:

driver = webdriver.Chrome()
driver.maximize_window() # now screen top-left corner == browser top-left corner 
driver.get("http://stackoverflow.com/questions")
question = driver.find_element_by_link_text("Questions")
y_relative_coord = question.location['y']
browser_navigation_panel_height = driver.execute_script('return window.outerHeight - window.innerHeight;')
y_absolute_coord = y_relative_coord + browser_navigation_panel_height
x_absolute_coord = question.location['x']

回答2:

There is no way to do this with 100% accuracy, but here is the best workaround that considers the browser window's offset, the toolbars in the window, and the scrolling position of the document:

# Assume there is equal amount of browser chrome on the left and right sides of the screen.
canvas_x_offset = driver.execute_script("return window.screenX + (window.outerWidth - window.innerWidth) / 2 - window.scrollX;")
# Assume all the browser chrome is on the top of the screen and none on the bottom.
canvas_y_offset = driver.execute_script("return window.screenY + (window.outerHeight - window.innerHeight) - window.scrollY;")
# Get the element center.
element_location = (element.rect["x"] + canvas_x_offset + element.rect["width"] / 2,
                    element.rect["y"] + canvas_y_offset + element.rect["height"] / 2)