Finding a string in an URL's source code with selenium webdriver - selenium

Question

I'm trying to extract a keyword/string from a website's source code using this python 2.7 script:
from selenium import webdriver
keyword = ['googleadservices']
driver = webdriver.Chrome(executable_path=r'C:\Users\Jacob\PycharmProjects\Testing\chromedriver_win32\chromedriver.exe')
driver.get('https://www.vacatures.nl/')
elem = driver.find_element_by_xpath("//*")
source_code = elem.get_attribute("outerHTML")
for searchstring in keyword:
if searchstring.lower() in str(source_code).lower():
print (searchstring, 'found')
else:
print (searchstring, 'not found')
The browser fortunately opens when the script is running, but I'm not able to extract the desired keywords from it's source code. Any help?

Answer 1

As others have said, the issue isn't your code but simply that googleadservice isn't present in the source code.
What I want to add, is that your code is a bit over engineered, since all you seem to do is either return true or false if a certain string is present in the source code.
You can achieve that much easier with a better xpath like //script[contains(text(),'googletagmanager')] and than use find_element_by_xpath and catch the possible NoSuchElementException. That might save you time and you don't need the for loop.
There are other possiblities as well, using ExpectedConditions or find_elements_by_xpath and then check if the returned list is greater than 0.

Answer 2

I observed that googleadservices is NOT present in the web page source code.
There is NO issue with the code.
I tried with GoogleAnalyticsObject, and it is found.
from selenium import webdriver
keyword = ['googleadservices', 'GoogleAnalyticsObject']
driver = webdriver.Chrome()
driver.get('https://www.vacatures.nl/')
elem = driver.find_element_by_xpath("//*")
source_code = elem.get_attribute("outerHTML")
for searchstring in keyword:
if searchstring.lower() in str(source_code).lower():
print (searchstring, 'found')
else:
print (searchstring, 'not found')
Instead of using //* to find the source code
elem = driver.find_element_by_xpath("//*")
source_code = elem.get_attribute("outerHTML")
Use the following code:
source_code = driver.page_source