Reputation: 13481
I´m trying to use argparse of Python but I cannot get a command line parameter.
Here my code:
DEFAULT_START_CONFIG='/tmp/config.json'
parser = argparse.ArgumentParser(description='Start the Cos service and broker for development purposes.')
parser.add_argument('-c', '--config', default=DEFAULT_START_CONFIG, action=FileAction, type=str, nargs='?',
help='start configuration json file (default:' + DEFAULT_START_CONFIG + ')')
args = parser.parse_args()
But then when I run my python script like:
./start.py -c /usr/local/config.json
Instead of getting this path it is getting the default value defined (/tmp/config.json
).
print args.config ---> "/tmp/config.json"
What I´m doing wrong here?
Upvotes: 0
Views: 656
Reputation: 17455
The standard documentation doesn't mention FileAction
. Instead there's a class FileType intended for type
argument, not for action
.
So I would write something like this:
DEFAULT_START_CONFIG='/tmp/config.json'
parser = argparse.ArgumentParser(description='Start the Cos service and broker for development purposes.')
parser.add_argument('-c', '--config', default=DEFAULT_START_CONFIG,
type=argparse.FileType('r'), help='start configuration json file')
args = parser.parse_args()
print(args)
This gives me the following:
$ python test3.py
Namespace(config=<open file '/tmp/config.json', mode 'r' at 0x7fd758148540>)
$ python test3.py -c
usage: test3.py [-h] [-c CONFIG]
test3.py: error: argument -c/--config: expected one argument
$ python test3.py -c some.json
usage: test3.py [-h] [-c CONFIG]
test3.py: error: argument -c/--config: can't open 'some.json': [Errno 2] No such file or directory: 'some.json'
$ touch existing.json
$ python test3.py -c existing.json
Namespace(config=<open file 'existing.json', mode 'r' at 0x7f93e27a0540>)
You may subclass argparse.FileType
to something like JsonROFileType
which would check if the supplied file is actually a JSON of expected format etc, but this seems to be out of the scope of the question.
Upvotes: 1