CODEWITHSUNDEEP
pythonseleniumselenium-webdriver
fightstarr20
fightstarr20

Reputation: 12608

Python Selenium continually check for phrase in page source

I am new to using Python. I am trying to write a script using Selenium that opens a Firefox browser and checks to see if a specific phrase is in the page source.

So far I have this...

from selenium import webdriver
driver = webdriver.Firefox()
driver.get('http://www.example.com/')
current_page_source = driver.page_source
print(current_page_source)

if "my phrase" in current_page_source :
    print ("Phrase Is Present")
else:
    print ("Phrase Is Not Present")

This is working great, but I would like this to continually check the source every 30 seconds.

Do I need a to use a for loop to achieve this?

Upvotes: 1

Views: 148

Answers (1)

mech
mech

Reputation: 2913

In order to run that code every 30 seconds, you need two things:

  1. You need to create a 30-second delay. The easiest way is to use the time module to create a delay. For example:

    import time
# some code
time.sleep(30) # delays for 30 seconds

  2. You need to repeat the action. The easiest way is to use a while loop that checks a condition. For example:

    keep_running = True

while(keep_running):
    # your repeatable code

    That way you can later set some logic in the code inside the while-loop that may set the keep_running variable to False, thus ending the loop (for example, once it's found the text you're looking for 1000 times, or if you press a button, etc).

Upvotes: 1

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