Question

在下面的示例中，我尝试设置以下命令行方案：

$ myapp.py flowers [-h]
$ myapp.py flowers plants [-h]
$ myapp.py flowers plants add [-h]

实际结果如下：

$ python myapp.py flowers
usage: myapp.py flowers [-h] {plants} ...
myapp.py flowers: error: too few arguments
$ python myapp.py flowers plants
usage: myapp.py flowers plants [-h] [--format {plain,json}] {add} ...
myapp.py flowers plants: error: too few arguments
$ python myapp.py flowers plants add
Adding flowers...  The input format is expected to be plain

问题是第二种情况。我打算将它作为一个隐含的“列表”操作。这该怎么做？

以下是我的示例的源代码。

import sys
import argparse

def _flowers_plants_list(args):
    print('if this were real, the output format would be %s' % (args.format))       
    print('Here are all the flowers:')
    print('flower 1: rose')
    print('flower 2: tulip')
    print('flower 3: daisy')

def _flowers_plants_add(args):
    print('Adding flowers...  The input format is expected to be %s' % (args.format))

def main():
    # myapp.py
    parser_main = argparse.ArgumentParser()
    subparsers_main = parser_main.add_subparsers()

    # myapp.py flowers [-h]
    parser_flowers = subparsers_main.add_parser("flowers")
    subparsers_flowers = parser_flowers.add_subparsers()

    # (this is supposed to be an implicit list - NOT WORKING)
    # myapp.py flowers plants [-h]
    parser_flowers_plants = subparsers_flowers.add_parser("plants")
    parser_flowers_plants.add_argument("--format", default='plain', choices=['plain', 'json'], help="what format should the result be in")
    parser_flowers_plants.set_defaults(func=_flowers_plants_list)
    subparsers_flowers_plants = parser_flowers_plants.add_subparsers()

    # myapp.py flowers plants add [-h]
    parser_flowers_plants_add = subparsers_flowers_plants.add_parser("add")
    parser_flowers_plants_add.add_argument("--format", default='plain', choices=['plain', 'json'], help="what format will the input be in")
    parser_flowers_plants_add.set_defaults(func=_flowers_plants_add)

    args = parser_main.parse_args()
    r = args.func(args)
    return r

if __name__ == "__main__":
    sys.exit(main())

Answer 1

在Py3中：

1420:~/mypy$ python3 stack50350537.py 
Traceback (most recent call last):
  File "stack50350537.py", line 40, in <module>
    sys.exit(main())
  File "stack50350537.py", line 36, in main
    r = args.func(args)
AttributeError: 'Namespace' object has no attribute 'func'
1421:~/mypy$ python3 stack50350537.py flowers
Traceback (most recent call last):
  File "stack50350537.py", line 40, in <module>
    sys.exit(main())
  File "stack50350537.py", line 36, in main
    r = args.func(args)
AttributeError: 'Namespace' object has no attribute 'func'
1421:~/mypy$ python3 stack50350537.py flowers plants
if this were real, the output format would be plain
Here are all the flowers:
flower 1: rose
flower 2: tulip
flower 3: daisy
1422:~/mypy$ python3 stack50350537.py flowers plants add
Adding flowers...  The input format is expected to be plain

在py2中，需要子分析符：

1423:~/mypy$ python2 stack50350537.py
usage: stack50350537.py [-h] {flowers} ...
stack50350537.py: error: too few arguments
1423:~/mypy$ python2 stack50350537.py flowers
usage: stack50350537.py flowers [-h] {plants} ...
stack50350537.py flowers: error: too few arguments
1423:~/mypy$ python2 stack50350537.py flowers plants
usage: stack50350537.py flowers plants [-h] [--format {plain,json}] {add} ...
stack50350537.py flowers plants: error: too few arguments
1423:~/mypy$ python2 stack50350537.py flowers plants add
Adding flowers...  The input format is expected to be plain

在Py3中，不需要子分段。最初这个更改是一个错误，这是因为required的测试方式发生了变化。但它已经足够长，现在它已成为一种功能。有一种方法可以让用户对此功能有更多的控制权，但我认为not required仍然是默认值。

Argparse with required subparser

如何使用argparse堆叠多个子命令并使用隐含的默认命令？

1 个答案: