I am opening a file which has 100,000 URL\'s. I need to send an HTTP request to each URL and print the status code. I am using Python 2.6, and so far looked at the many con
I know this is an old question, but in Python 3.7 you can do this using asyncio and aiohttp.
import asyncio
import aiohttp
from aiohttp import ClientSession, ClientConnectorError
async def fetch_html(url: str, session: ClientSession, **kwargs) -> tuple:
try:
resp = await session.request(method="GET", url=url, **kwargs)
except ClientConnectorError:
return (url, 404)
return (url, resp.status)
async def make_requests(urls: set, **kwargs) -> None:
async with ClientSession() as session:
tasks = []
for url in urls:
tasks.append(
fetch_html(url=url, session=session, **kwargs)
)
results = await asyncio.gather(*tasks)
for result in results:
print(f'{result[1]} - {str(result[0])}')
if __name__ == "__main__":
import pathlib
import sys
assert sys.version_info >= (3, 7), "Script requires Python 3.7+."
here = pathlib.Path(__file__).parent
with open(here.joinpath("urls.txt")) as infile:
urls = set(map(str.strip, infile))
asyncio.run(make_requests(urls=urls))
You can read more about it and see an example here.