我希望Python argparse抛出异常而不是用法

Question

I don't think this is possible, but I want to handle exceptions from argparse myself.

For example:

import argparse
parser = argparse.ArgumentParser()
parser.add_argument('--foo', help='foo help', required=True)
try:
    args = parser.parse_args()
except:
    do_something()

When I run it:

$ myapp.py
usage: myapp --foo foo
myapp: error: argument --foo is required

But I want it to fall into the exception instead.

Answer 1

You can subclass ArgumentParser and override the error method to do something different when an error occurs:

class ArgumentParserError(Exception): pass

class ThrowingArgumentParser(argparse.ArgumentParser):
    def error(self, message):
        raise ArgumentParserError(message)

parser = ThrowingArgumentParser()
parser.add_argument(...)
...

Answer 2

in my case, argparse prints 'too few arguments' then quit. after reading the argparse code, I found it simply calls sys.exit() after printing some message. as sys.exit() does nothing but throws a SystemExit exception, you can just capture this exception.

so try this to see if it works for you.

    try:
        args = parser.parse_args(args)
    except SystemExit:
        .... your handler here ...
        return