Python Selenium Check if .click() was clickable or not-CodePudding

I would like to use an if function that tells me whether a button has been clicked or not in my code given below, if yes then "a" should be output if no then "b".

element = driver.find_element(By.XPATH, '//*[contains(concat( " ", @class, " " ), concat( " ", "css-xf3ahq", " " ))][text()="Button"]')
element.click()

if element == True:
    print("a")

else:
    print("b")

up to now b has always been written although the button is clickable

CodePudding user response：

Selenium has: element_to_be_clickable() which may be a better solution to what you're trying to do. i.e.

from selenium.webdriver.support import expected_conditions as EC
try:
    WebDriverWait(driver, 10).until(EC.element_to_be_clickable(By.XPATH, '//*[contains(concat( " ", @class, " " ), concat( " ", "css-xf3ahq", " " ))][text()="Button"]'))
    print('a')

except selenium.common.exceptions.TimeoutException:
    print('b')

CodePudding user response：

you can use element_to_be_clicked condition

like this:

element = WebDriverWait(driver, 10).until(
EC.element_to_be_clickable((By.ID, "myDynamicElement"))
)

CodePudding user response：

try:
    element = driver.find_element(By.XPATH, '//*[contains(concat( " ", @class, " " ), concat( " ", "css-xf3ahq", " " ))][text()="Button"]')
    element.click()
    print('Clickable')
except:
    print('Not clickable')

Just try to click on it otherwise print error if it isn't. You can check for presence with a webdriver and handle different errors with multiple excepts if you want. Not sure if you can do it with an if here.