Python get header status code of url's from an import csv list and print

Question

Im new to Python and trying to import a list of urls from a csv and check the header status of each. So far ive got the following which prints the list of urls:

import csv
from requests_html import HTMLSession
from bs4 import BeautifulSoup
import requests


with open('list.csv', 'r') as f:
    reader = csv.reader(f)
    for row in reader:
        url = " ".join(row)
        print(url)

I now want to print the header status code next to each url, so ive tried this which doesnt work. Any help would be appreciated:

with open('list.csv', 'r') as f:
    reader = csv.reader(f)
    for row in reader:
        url = " ".join(row)
        headers = requests.head(url.get('href'))
        print(url, (headers.status_code))

The error i get is:

headers = requests.head(url.get('href')) AttributeError: 'str' object has no attribute 'get'

I need to output something like:

https://www.domainone.com 200
https://www.domaintwo.com 200
https://www.domainthree.com 404

What doesn't work exactly? Do you see any errors, if yes it will really help to put that in your post. — AKS
– AKS, Commented Mar 17, 2020 at 16:41

Yun Luo · Accepted Answer · 2020-03-17 16:49:05Z

1

you already have the url and the response. just print them out:

resp = requests.head(url)
print('%s %d' % (url, resp.status_code))

answered Mar 17, 2020 at 16:49

Yun Luo

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