how to download videos from a site using python
recently I`m watching some videos of CS 162 from this site
and when I tried to find url of video I found out there is no url in inspect elements tab.
but I found the url of video in network tab. you can see this tab near inspect elements of site in opera.
I tried to get data of this tab using "requests" but It didn't turn out to be a clear url I was looking for.
Is there a way to gain data of network tab using python?
check this site if you can't find network tab.
thanks;
python web-scraping
add a comment |
recently I`m watching some videos of CS 162 from this site
and when I tried to find url of video I found out there is no url in inspect elements tab.
but I found the url of video in network tab. you can see this tab near inspect elements of site in opera.
I tried to get data of this tab using "requests" but It didn't turn out to be a clear url I was looking for.
Is there a way to gain data of network tab using python?
check this site if you can't find network tab.
thanks;
python web-scraping
1
r = requests.get(url, stream=True)
– N.S
Jan 17 at 21:56
add a comment |
recently I`m watching some videos of CS 162 from this site
and when I tried to find url of video I found out there is no url in inspect elements tab.
but I found the url of video in network tab. you can see this tab near inspect elements of site in opera.
I tried to get data of this tab using "requests" but It didn't turn out to be a clear url I was looking for.
Is there a way to gain data of network tab using python?
check this site if you can't find network tab.
thanks;
python web-scraping
recently I`m watching some videos of CS 162 from this site
and when I tried to find url of video I found out there is no url in inspect elements tab.
but I found the url of video in network tab. you can see this tab near inspect elements of site in opera.
I tried to get data of this tab using "requests" but It didn't turn out to be a clear url I was looking for.
Is there a way to gain data of network tab using python?
check this site if you can't find network tab.
thanks;
python web-scraping
python web-scraping
asked Jan 17 at 21:55
N.SN.S
300312
300312
1
r = requests.get(url, stream=True)
– N.S
Jan 17 at 21:56
add a comment |
1
r = requests.get(url, stream=True)
– N.S
Jan 17 at 21:56
1
1
r = requests.get(url, stream=True)
– N.S
Jan 17 at 21:56
r = requests.get(url, stream=True)
– N.S
Jan 17 at 21:56
add a comment |
1 Answer
1
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oldest
votes
I write this code base on your comment:
import requests
url = 'your site'
r = requests.get(url, stream=True) # your comment
print(r)
in the output there is a <iframe> tag which has a src of a .mp4.
I think you should grab that using bs4 and then download the video you want.
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
I write this code base on your comment:
import requests
url = 'your site'
r = requests.get(url, stream=True) # your comment
print(r)
in the output there is a <iframe> tag which has a src of a .mp4.
I think you should grab that using bs4 and then download the video you want.
add a comment |
I write this code base on your comment:
import requests
url = 'your site'
r = requests.get(url, stream=True) # your comment
print(r)
in the output there is a <iframe> tag which has a src of a .mp4.
I think you should grab that using bs4 and then download the video you want.
add a comment |
I write this code base on your comment:
import requests
url = 'your site'
r = requests.get(url, stream=True) # your comment
print(r)
in the output there is a <iframe> tag which has a src of a .mp4.
I think you should grab that using bs4 and then download the video you want.
I write this code base on your comment:
import requests
url = 'your site'
r = requests.get(url, stream=True) # your comment
print(r)
in the output there is a <iframe> tag which has a src of a .mp4.
I think you should grab that using bs4 and then download the video you want.
edited Jan 18 at 18:49
Infected Drake
703418
703418
answered Jan 17 at 22:24
roserose
312
312
add a comment |
add a comment |
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r = requests.get(url, stream=True)
– N.S
Jan 17 at 21:56