Program to check Website's URLs Status with Python.

The logic of the Python code below is really simple:

- First you need pip install advertools which will convert your sitemap to dataframe.
- Then import all necessary libraries.
- Convert sitemap to DF and then to list.
- Slice the list if you want to check just part of your website, e.g. just first 100 URLs.
- Using FOR loop check the status code of each URL, response 200 is oK.

Checking Website's URLs Status Python code:

import advertools as adv
import pandas as pd
import urllib.request
import time

sitemap_urls = adv.sitemap_to_df("https://python-code.pro/static/img/sitemap.xml")

url = sitemap_urls["loc"].to_list()
url=url[:5]

for i in url:
          status_code = urllib.request.urlopen(i).getcode()
          print(i)
          print(status_code)
          time.sleep(3)

print(f'Totally {len(url)} URLs checked, status code 200 is oK.')
print('Thats all, folks !!!')

OUT: INFO:root:Getting https://python-code.pro/static/img/sitemap.xml
https://python-code.pro/
200
https://python-code.pro/robots.txt
200
https://python-code.pro/regression-models-python-cheatsheets/
200
https://python-code.pro/regression-models-r-cheatsheets/
200
https://python-code.pro/classification-models-python-r-cheatsheets/
200
Totally 5 URLs checked, status code 200 is oK.
Thats all, folks !!!