How to throw error and exit with a custom message in python

How to throw error and exit with a custom message in python

Calling sys.exit with a string will work. The docs mention this use explicitly:

In particular, sys.exit(some error message) is a quick way to exit a program when an error occurs.

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 dont 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)

How to throw error and exit with a custom message in python

I know this is an old thread, however you can also raise an error like this:

raise SystemExit(Error: 3 processes cannot run simultaneously.)

One advantage of this approach is that you dont have to import the Python sys module. This works on Linux with Python 3 and Python 2. I have not tested it on Windows or Mac OS.

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