How to optimise the code evaluation in Selenium with Python

Question

I would like to extract the span text located within the class SOME_CLASS_NAME. The class SOME_CLASS_NAME might not always available. Also, it take some time before the class properly visible if it does exist.

To handle this, the following code were propsed.

Dirty workaround to check whether the class available or not

size_len = WebDriverWait(self.browser, 20).until(EC.presence_of_element_located(
                    (By.XPATH, './/span[@class = "SOME_CLASS_NAME"]')))

OR

 if len( self.browser.find_elements_by_xpath('.//span[@class = "SOME_CLASS_NAME"]') ) > 0 :  
        element =self.browser.find_elements_by_xpath( './/span[@class = "SOME_CLASS_NAME"]' )[ 0 ].text
    
 else:
         element = '0'

While the above approach work, but,I found it very inefficient to evaluate the class for 3 times.

May I know if there is a way to make the above code more efficient?

KunduK · Accepted Answer · 2020-04-27 09:54:22Z

1

Use try..except block and use visibility_of_element_located() instead presence_of_element_located()

try:
    print(WebDriverWait(self.browser, 10).until(EC.visibility_of_element_located((By.XPATH, './/span[@class = "SOME_CLASS_NAME"]'))).text)
except:
    print("element no available")

answered Apr 27, 2020 at 9:54

KunduK

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2 Comments

rpb Over a year ago

Thanks for the reply. Should the TimeoutException: be there? or this is some special case?

KunduK Over a year ago

@balandongiv : You can use ` except TimeoutException:` If you want but with that you can't continue further.if you want continue to the next step then use except.

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