Answers for "raise for status python"

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
404

python 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")