CODEWITHSUNDEEP
pythonpython-requests
question.it
question.it

Reputation: 2968

How to handle timeout error in request.head(url)?

Script:

>>> import requests
>>> print(requests.head('https://knoema.com/BISDPPLS2016/bis-property-prices-long-series'))
<Response [200]>

>>> print(requests.head('https://beta.knoema.org/BISDPPLS2016/bis-property-prices-long-series'))
Connection to beta.knoema.org timed out.

I tried below as well, but same error:

>>> print(requests.head('https://beta.knoema.org/IRFCL2016Jul', timeout=5))

I want to stop requests.head(url) if it does not exist or is not loading for any reason. How to achieve this?

Upvotes: 2

Views: 4297

Answers (1)

Arpit Solanki
Arpit Solanki

Reputation: 9931

You have to handle exception using try-catch

try: 
    requests.head('https://beta.knoema.org/BISDPPLS2016/bis-property-prices-long-series')
except requests.exceptions.Timeout as e:
    print e

Also you can set a large timeout to avoid error due to slow connection

requests.head('https://beta.knoema.org/BISDPPLS2016/bis-property-prices-long-series', \
               timeout=5)

Here timeout is 5 seconds

Upvotes: 4

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