Detect if url represents streaming link - python-requests

Question

I am trying to indicate if a URL contains streaming.
I want to close the connection if the link is a streaming link.
Is there a better way to check if the link is a streaming link?
Can I do the request without using stream=True? Because I am only interested in content-Type:'text/*'
What did I try:
I tried to send the header Connection: close but it didn't stop the stream.
I also tried to send the header Connection: Keep-Alive and Keep-Alive: timeout=5, max=1 didn't do the job.
I tried to pass stream=False but I couldn't receive request.text.
My attempt:
def ignore_stream():
with requests.get('https://google.com', timeout=5, stream=True) as r:
if 'Content-Type' in r.headers:
if 'audio' in r.headers['Content-Type'] or 'video' in r.headers['Content-Type']:
print(r.headers['Content-Type'])
print('streaming link .. close the connection')
r.close()
else:
print('text page .. get text content')
print(r.text)
if __name__ == '__main__':
ignore_stream()
The code above does close the stream connection, but I am afraid, that won't be enough for stream detection.

Answer 1

Your connection gets closed because of that with statement, as it is calling Response.close() automatically at the end of the statement. There is no need to do r.close(), just leave the statement do the job or remove it and do it yourself.
Sending the request with the stream flag as True will only download the response headers at first (more info here). This means that if you only want the text/** responses (and mark the rest as "streaming" URLs), you can achieve it at this point. See the demo below and let me know if it suits you:
# using "with" statement
def ignore_stream_with():
with requests.get('https://google.com', timeout=5, stream=True) as r:
if 'Content-Type' in r.headers:
if 'audio' in r.headers['Content-Type'] or 'video' in r.headers['Content-Type']:
print(r.headers['Content-Type'])
print('Link detected as streaming... Leaving "with" to close the connection.')
else:
print('Link detected as text!')
print('Response content:')
print(r.text)
# do more stuff with your response...
pass
# NOT using "with" statement
def ignore_stream_without():
r = requests.get('https://google.com', timeout=5, stream=True)
if 'Content-Type' in r.headers:
if 'audio' in r.headers['Content-Type'] or 'video' in r.headers['Content-Type']:
print(r.headers['Content-Type'])
print('Link detected as streaming... Leaving "with" to close the connection.')
else:
print('Link detected as text!')
print('Response content:')
print(r.text)
# do more stuff with your response...
pass
# remember always to close the connection!
r.close()
if __name__ == '__main__':
ignore_stream()