get request python
import requests
x = requests.get('https://w3schools.com')
print(x.status_code)get request python
import requests
x = requests.get('https://w3schools.com')
print(x.status_code)python print return code of requests
>>> import requests
>>> r = requests.get('http://httpbin.org/status/404')
>>> r.status_code
404python raise error exit
# There are 3 approaches, the first as lvc mentioned is using sys.exit
sys.exit('My error message')
# The second way is using print, print can write almost anything including an error message
print >>sys.stderr, "fatal error" # Python 2.x
print("fatal error", file=sys.stderr) # Python 3.x
# The third way is to rise an exception which I don't like because it can be try-catch
raise SystemExit('error in code want to exit')
# it can be ignored like this
try:
raise SystemExit('error in code want to exit')
except:
print("program is still open")Copyright © 2021 Codeinu
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