Request (or httplib2.request) taking too long to return status code

Question

I'm trying to check if some file links are up or not. I've tried with requests library:

import requests
url = 'http://releases.libreelec.tv/LibreELEC-RPi2.arm-9.0.1.img.gz'
print(requests.get(url).status_code)

And also with httplib2:

import httplib2
url = 'http://releases.libreelec.tv/LibreELEC-RPi2.arm-9.0.1.img.gz'
http = httplib2.Http()
resp = http.request(url)[0]
print(resp.status)

As the link is good.. it takes too long (I've never waited it to finish) to return a '200' status code. It works fine for forbidden (403) code. The point is i need it to be fast.

score 1 · Answer 1 · answered Mar 01 '19 at 08:17

1

If you need to get response code only, you can use requests.head:

import requests
try:
    r = requests.head('http://releases.libreelec.tv/LibreELEC-RPi2.arm-9.0.1.img.gz')
    print(r.status_code)
except:
    print('problem')

As shown in this topic, it should be faster as it gets only head, however please check if it is fast enough for your purpose.

answered Mar 01 '19 at 08:17

Daweo

31,313
3
12
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for the specific situation, I've tried 'head' before.. it returns 200 for 403 status codes, for some reason. – Tiago Aquino Mar 01 '19 at 09:53

score 0 · Accepted Answer · answered Mar 01 '19 at 08:25

0

fast:

r = requests.get(url, stream=True)

ref:python requests is slow

answered Mar 01 '19 at 08:25

Vox

506
2
13

Request (or httplib2.request) taking too long to return status code

2 Answers2