docopt +模式验证

Question

是否有更好的方法来处理此验证：

#!/usr/bin/env python
""" command.

Usage:
  command start ID
  command finish ID FILE
  command (-h | --help)
  command (-v | --version)

Arguments:
  FILE     input file
  PATH     out directory

Options:
  -h --help     Show this screen.
  -v --version  Show version.
"""

from docopt import docopt
from schema import Schema, Use, SchemaError

if __name__ == '__main__':
    args = docopt(__doc__, version='command alpha')

    # Remove False or None keys from args dict
    for k, v in args.items():
        if (not v):
            args.pop(k)

    if 'start' in args:
        args.pop('start')
        schema = Schema({
            'FILE': Use(open, error='FILE should be readable'),
            'ID': Use(int, error='ID should be an int'),
        })
    elif 'finish' in args:
        args.pop('finish')
        schema = Schema({
            'FILE': Use(open, error='FILE should be readable'),
            'ID': Use(int, error='ID should be an int'),
        })

    try:
        args = schema.validate(args)
    except SchemaError as e:
        exit(e)

    print(args)

Answer 1

我会做以下事情：

#!/usr/bin/env python
"""Command.

Usage:
  command start ID
  command finish ID FILE
  command (-h | --help)
  command (-v | --version)

Arguments:
  ID
  FILE     input file

Options:
  -h --help     Show this screen.
  -v --version  Show version.

"""
from docopt import docopt
from schema import Schema, Use, Or, SchemaError

if __name__ == '__main__':
    args = docopt(__doc__, version='command alpha')

    id_schema = Use(int, error='ID should be an int')
    file_schema = Or(None, Use(open, error='FILE should be readable'))
    try:
        args['ID'] = id_schema.validate(args['ID'])
        args['FILE'] = file_schema.validate(args['FILE'])
    except SchemaError as e:
        exit(e)

    print(args)

虽然我希望架构可以使用单个架构表达相同的内容，而不是两个。我将尝试在将来制作类似以下的模式：

schema = Schema({'ID': Use(int, error='ID should be an int'),
                 'FILE': Or(None, Use(open, error='FILE should be readable')),
                 object: object})

object: object表示我只关心'ID'和'FILE'，并且所有其他键/值都可以是任意对象。

更新

从版本0.2.0开始，架构现在可以正确处理此案例：

schema = Schema({'ID': Use(int, error='ID should be an int'),
                 'FILE': Or(None, Use(open, error='FILE should be readable')),
                 object: object})