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How to click a button using selenium and draw on a canvas in python?

so I am trying to automate a program that would log in to a google account I created, go to this canvas website, and draw a circle,
(Circle just as a placeholder because im trying to make it draw some cool stuff like a car, just to test if it will work first.) But the main issue is when you first go into the website, there is a pop up that displays and that pop up has 2 options, "learn more" and "Get started". I tried to make selenium click on the "Get started" using driver.find_element_by_id('Get-started').click but it does not seem to work, I also tried to use CSS selector but that does not seem to work either. So, I'm stuck on this. Any advice or help to click the get started button? (Feel free to also give me advice on how to draw in the canvas as well!)
Here's the HTML:
<paper-button id="get-started" dialog-confirm="" class="primary" aria-label="Get started" role="button" tabindex="0" animated="" elevation="0" aria-disabled="false">
Get started
</paper-button>
here's the code:
from selenium import webdriver
import time
from PrivateInfo import*
driver = webdriver.Chrome(r"D:\chromeDriver\chromedriver.exe")
link = "https://www.google.com/"
driver.get(link)
def linkText(element):
button = driver.find_element_by_link_text(element)
button.click()
def byClass(className):
button2 = driver.find_element_by_class_name(className)
button2.click()
def type(text, elements):
input = driver.find_elements_by_name(elements)
for element in input:
driver.implicitly_wait(2)
element.send_keys(text)
linkText("Sign in")
type(email, "identifier")
byClass("VfPpkd-vQzf8d")
type(pw, "password")
driver.find_element_by_id("passwordNext").click()
time.sleep(1)
driver.get("https://canvas.apps.chrome/")
driver.implicitly_wait(3)
driver.find_element_by_css_selector('paper-button[id="get-started"]').click()
edit: I also tried this
getStart = driver.find_elements_by_css_selector('paper-button[id="get-started"]')
for start in getStart:
start.click()
it doesn't give me any errors but it does not do anything.
Ah yeah, forgot to mention but im new to using selenium.

The content of the popup is nested inside shadowRoots.
More on ShadowRoot here:
The ShadowRoot interface of the Shadow DOM API is the root node of a
DOM subtree that is rendered separately from a document's main DOM
tree.
To be able to control the element you will need to switch between DOMs. Here's how:
drawing_app_el = driver.execute_script("return arguments[0].shadowRoot", driver.find_element(By.CSS_SELECTOR, 'drawing-app'))
This code will retrieve the drawing_app first and then return the content of the shadowRoot.
To have access to the button getStarted, here's how:
# Get the shadowRoot of drawing-app
drawing_app_el = driver.execute_script("return arguments[0].shadowRoot", driver.find_element(By.CSS_SELECTOR, 'drawing-app'))
# Get the shadowRoot of welcome-dialog within drawing_app
welcome_dialog_el = driver.execute_script("return arguments[0].shadowRoot", drawing_app_el.find_element(By.CSS_SELECTOR, 'welcome-dialog'))
# Get the paper-dialog from welcome_dialog
paper_dialog_el = welcome_dialog_el.find_element(By.CSS_SELECTOR, 'paper-dialog')
# Finally, retrieve the getStarted button
get_started_button = paper_dialog_el.find_element(By.CSS_SELECTOR, '#get-started')
get_started_button.click()

How to click on like/dislike button of YouTube video using selenium webdriver python?

my code :
react_on_video = (input("React: ")).upper()
video_link = "https://www.youtube.com/watch?v=blablabla"
(After login with my gmail account)
driver.get(video_link)
driver.execute_script("window.scrollTo(0, 300);")
time.sleep(1)
if react_on_video == "LIKE":
Like_button_Full_xpath = "/html/body/ytd-app/div/ytd-page-manager/ytd-watch-
flexy/div[5]/div[1]/div/div[8]/div[2]/ytd-video-primary-info-
renderer/div/div/div[3]/div/ytd-menu-renderer/div[1]/ytd-toggle-button-renderer[1]/a/yt-
icon-button/button/yt-icon"
driver.find_element_by_xpath(Like_button_Full_xpath).click()
if react_on_video == "DISLIKE":
Dislike_button_Full_xpath = "/html/body/ytd-app/div/ytd-page-manager/ytd-watch-
flexy/div[5]/div[1]/div/div[8]/div[2]/ytd-video-primary-info-
renderer/div/div/div[3]/div/ytd-menu-renderer/div[1]/ytd-toggle-button-renderer[2]/a/yt-
icon-button/button/yt-icon"
driver.find_element_by_xpath(Dislike_button_Full_xpath).click()
But every time it shows No such element exception!
[selenium.common.exceptions.NoSuchElementException: Message: no such element:]
I've tried with other selectors and xpaths; but none of them worked! Is there any permanent solution? Because once I was able to click on the buttons with this xpath. I had to change xpath multiple times.
Like_button_xpath : '//[#id="top-level-buttons"]/ytd-toggle-button-renderer1/a'
Dislike_button_xpath : '//[#id="top-level-buttons"]/ytd-toggle-button-renderer[2]/a'

I found Solution. Try xPath
Like_button_Full_xpath = "//div[3]/div/ytd-menu-renderer/div/ytd-toggle-button-renderer/a/yt-icon-button/button/yt-icon"
Tell me if it work.

Instead of using xpath, I would use a css selector. I can't guarantee that this selector will be unique without seeing the full HTML, but I would try finding the "Like" button like this:
driver.find_element_by_css_selector('yt-icon.style-scope.ytd-toggle-button-renderer')

Python selenium can't click a button execute script

I am trying to click a button from the Linkedin public page feed
https://www.linkedin.com/company/bbc-news/
and the button is a post's toolbar button with 3 dots
here is the sample screenshot
and my code is
self.browser.execute_script("document.getElementsByClassName('feed-shared-control-menu__trigger artdeco-button artdeco-button--tertiary artdeco-button--muted artdeco-button--1 artdeco-button--circle artdeco-dropdown__trigger artdeco-dropdown__trigger--placement-bottom ember-view')[1].scrollIntoView();")
self.browser.execute_script("document.getElementsByClassName('feed-shared-control-menu__trigger artdeco-button artdeco-button--tertiary artdeco-button--muted artdeco-button--1 artdeco-button--circle artdeco-dropdown__trigger artdeco-dropdown__trigger--placement-bottom ember-view')[1].click();")
it doesn't return any errors
please kindly help me solve this issue
Thanks in advance for any help
Note: I have already tried webDriverwait , find_elements_by_class_name

The element has an id associated to it, so you can use that id to get the xpath of the element and then you can click it. Also, it is recommended to use the selenium click instead of the javascript executor click(which you are currently using).
Your code should be like:
self.browser.find_element_by_xpath("//button[#id='ember160']//li-icon").click()
If you want to use javascript executor click, then your code should be like:
element = self.browser.find_element_by_xpath("//button[#id='ember160']//li-icon")
driver.execute_script("arguments[0].click();", element)

Not able to click a button in Selenium (Booking.com)

I am programming a python scraper with help of Selenium. The first few steps are:
goes on booking.com, insert a city name, selects the first date and then tries to open the check-out calendar.
Here is where my problem occurs. I am not able to click the check-out calendar button (The important are of the website).
I tried to click every element regarding to the to check-out calendar (The elements of check-out calendar) with element.click(). I also tried the method
element = self.browser.find_element_by_xpath('(//div[contains(#class,"checkout-field")]//button[#aria-label="Open calendar"])[1]') self.browser.execute_script("arguments[0].click();", element)
It either does nothing (in case of execute.script() and click() on div elements) or it throws following exception when directly clicking the button:
Element <button class="sb-date-field__icon sb-date-field__icon-btn bk-svg-wrapper"
type="button"> is not clickable at point (367.5,316.29998779296875)
because another element <div class="sb-date-field__display"> obscures it
Here is a short code to test it:
browser = webdriver.Firefox()
browser.get("https://www.booking.com/")
wait = WebDriverWait(browser, 5)
element = wait.until(EC.presence_of_element_located((
By.XPATH, '(//div[contains(#class,"checkout-field")]//button[#aria-label="Open calendar"])[1]')))
element = wait.until(EC.element_to_be_clickable((
By.XPATH, '(//div[contains(#class,"checkout-field")]//button[#aria-label="Open calendar"])[1]')))
element.click()
I have a temporarily solution for my problem but I am not satisfied with it.
element = browser.find_element_by_xpath('(//div[contains(#class,"checkout-field")]//button[#aria-label="Open calendar"])[1]')
hov = ActionChains(browser).move_to_element(element)
hov.click().perform()
This will open the calendar by hovering over the object and clicking it. This strangely opens the calendar.
The methods mentioned above still don't work.

Define clicka as an xpath. Now use executescript to click the element.
driver.execute_script("arguments[0].click();", clicka)

I'm not 100% sure that I got everything you posted, because the layout is a bit messy.
However, I tried to test the issue with both Selenium Java and Firefox Scratchpad (a Web Developer tool that allows to run JavaScript scripts) and it worked perfectly - the button was clickable on both of them.
If you're interested in further testing using this tool, this is the code I've used:
In JavaScript:
function getElementByXpath(path) {
return document.evaluate(path, document, null, XPathResult.FIRST_ORDERED_NODE_TYPE, null).singleNodeValue;
}
var myElement = getElementByXpath('(//div[contains(#class,"checkout-field")]//button[#aria-label="Open calendar"])[1]')
myElement.click()
and in Java:
FirefoxDriver driver = new FirefoxDriver();
WebDriverWait wait = new WebDriverWait(driver, 10);
driver.navigate().to("https://www.booking.com");
wait.until(ExpectedConditions.elementToBeClickable(By.xpath("(//div[contains(#class,'checkout-field')]//button[#aria-label='Open calendar'])[1]")));
driver.findElement(By.xpath("(//div[contains(#class,'checkout-field')]//button[#aria-label='Open calendar'])[1]")).click();
System.out.println("success");

if your are having the control check out button on all the web site managing with explicit wait required lots of codding you can use implicit wait below is in the java.
System.setProperty("webdriver.chrome.driver",
"G:\\TopsAssignment\\SampleJavaExample\\lib\\chromedriver.exe");
WebDriver driver = new ChromeDriver();
driver.manage().window().maximize();
driver.manage().timeouts().implicitlyWait(5, TimeUnit.SECONDS);

Python+Selenium, can't click the 'button' wrapped by span

I am new to selenium here. I am trying to use selenium to click a 'more' button to expand the review section everytime after refreshing the page.
The website is TripAdvisor. The logic of more button is, as long as you click on the first more button, it will automatically expand all the review sections for you. In other words, you just need to click on the first 'more' button.
All buttons have a similar class name. An example is like taLnk.hvrIE6.tr415411081.moreLink.ulBlueLinks. Only the numbers part changes everytime.
The full element look like this:
<span class="taLnk hvrIE6 tr413756996 moreLink ulBlueLinks" onclick=" var options = {
flow: 'CORE_COMBINED',
pid: 39415,
onSuccess: function() { ta.util.cookie.setPIDCookie(2247); ta.call('ta.servlet.Reviews.expandReviews', {type: 'dummy'}, ta.id('review_413756996'), 'review_413756996', '1', 2247);; window.location.hash = 'review_413756996'; }
};
ta.call('ta.registration.RegOverlay.show', {type: 'dummy'}, ta.id('review_413756996'), options);
return false;
">
More </span>
I have tried several ways to get the button click. But since it is an onclick event wrapped by span, I can't successfully get it clicked.
My last version looks like this:
driver = webdriver.Firefox()
driver.get(newurl)
page_source = driver.page_source
soup = BeautifulSoup(page_source)
moreID = soup.find("span", class_=re.compile(r'.*\bmoreLink\b.*'))['class']
moreID = '.'.join(moreID[0:(len(moreID)+1)])
moreButton = 'span.' + moreID
button = driver.find_element_by_css_selector(moreButton)
button.click()
time.sleep(10)
However, I keep getting the error message like this:
WebDriverException: Message: Element is not clickable at point (318.5,
7.100006103515625). Other element would receive the click....
Can you advise me on how to fix the problem? Any help will be appreciated!

WebDriverException: Message: Element is not clickable at point (318.5, 7.100006103515625). Other element would receive the click....
This error to be occur when element is not in the view port and selenium couldn't click due to some other overlay element on it. In this case you should try one of these following solution :-
You can try using ActionChains to reach that element before click as below :-
from selenium.webdriver.common.action_chains import ActionChains
button = driver.find_element_by_css_selector(moreButton)
ActionChains(button).move_to_element(element).click().perform()
You can try using execute_script() to reach that element before click as :-
driver.execute_script("arguments[0].scrollIntoView(true)", button)
button.click()
You can try using JavaScript::click() with execute_script() but this JavaScript::click() defeats the purpose of the test. First because it doesn't generate all the events like a real click (focus, blur, mousedown, mouseup...) and second because it doesn't guarantee that a real user can interact with the element. But to get rid from this issues you can consider it as an alternate solution.
driver.execute_script("arguments[0].click()", button)
Note:- Before using these options make sure you're trying to interact with correct element using with correct locator, otherwise WebElement.click() would work well after wait until element visible and clickable using WebDriverWait.

Try using an ActionChains:
from selenium.webdriver.common.action_chains import ActionChains
# Your existing code here
# Minus the `button.click()` line
ActionChains(driver).move_to_element(button).cli‌​ck().perform()
I have used this technique when I need to click on a <div> or a <span> element, rather than an actual button or link.