如何打印Argparse主解析器的帮助(用法)部分而不是subparser的用法?

时间:2017-10-15 12:07:44

标签: python-2.7 argparse

我一直在用Argparse编写命令行程序已经有一段时间了,我试图用这样的方式编写它,当用户将以下内容提供给命令行时:

$python my_script.py -h

将打印出帮助部分(用法),打印出主解析器的帮助部分,以及子分析器的简要概述。

但是现在,无论何时我输入上一行到我的终端,我都没有收到任何用法,而是获得了大量的追溯和以下错误:

TypeError: expected string or buffer

在使用基于argparse的命令行程序之前,我从未发生过这个错误。此外,如果我提供其中一个子分析器的名称,

$python my_script.py subparserA -h

我打印出subparser的用法。其他子分析师也是如此。

那么为什么我不可能获得主解析器的用法?这之前对我有用,所以我不知道为什么它现在不起作用。我真的希望用户能够查看可用的不同子分析器的概述。

我的基本代码目前按以下方式设置:

import argparse
import sys

if __name__ == "__main__":
    Parser = argparse.ArgumentParser(prog= "My_program")

    Parser.description= "This program does A and B things."
    subparsers= Parser.add_subparsers(help= "SubparserA does A things and SubparserB does B things", dest='mode')

    subparserA= subparsers.add_parser("subparserA", help= "Additional explanation of what A things entail")

    subparserA.add_arguments("-foo", required=True, help= "foo is needed for SubparserA to work")

    subparserB= subparsers.add_parser("subparserB", help="Additional explanation of what B things entail")

    subparserB.add_argument("-bar", required=True, help= "bar is needed for SubparserB to work")

    args= Parser.parse_args()

    if args.mode == "subparserA":
        ###do things pertinent to subparserA
    elif args.mode== "subparserB":
        ###do things pertinent to subparserB
    else:
        argparse.print_help()
        argparse.ArgumentError("too few arguments")

更新

以下是错误的完整回溯:

Traceback (most recent call last):
  File "my_program.py", line 164, in <module>
    args= Parser.parse_args()
  File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/argparse.py", line 1701, in parse_args
    args, argv = self.parse_known_args(args, namespace)
  File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/argparse.py", line 1733, in parse_known_args
    namespace, args = self._parse_known_args(args, namespace)
  File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/argparse.py", line 1939, in _parse_known_args
    start_index = consume_optional(start_index)
  File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/argparse.py", line 1879, in consume_optional
    take_action(action, args, option_string)
  File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/argparse.py", line 1807, in take_action
    action(self, namespace, argument_values, option_string)
  File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/argparse.py", line 996, in __call__
    parser.print_help()
  File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/argparse.py", line 2340, in print_help
    self._print_message(self.format_help(), file)
  File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/argparse.py", line 2314, in format_help
    return formatter.format_help()
  File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/argparse.py", line 281, in format_help
    help = self._root_section.format_help()
  File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/argparse.py", line 211, in format_help
    func(*args)
  File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/argparse.py", line 485, in _format_text
    return self._fill_text(text, text_width, indent) + '\n\n'
  File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/argparse.py", line 621, in _fill_text
    text = self._whitespace_matcher.sub(' ', text).strip()
TypeError: expected string or buffer

2 个答案:

答案 0 :(得分:0)

您应该使用

Parser.print_help()
Parser.error('too few arguments')

这是使用现有Parser对象的方法。

当我运行你的脚本时

1019:~/mypy$ python stack46754855.py 
Traceback (most recent call last):
  File "stack46754855.py", line 10, in <module>
    subparserA= subparsers.add_parser("subparserA", help= "Additional explanation of what A things entail", dest= 'mode')
  File "/usr/lib/python2.7/argparse.py", line 1066, in add_parser
    parser = self._parser_class(**kwargs)
TypeError: __init__() got an unexpected keyword argument 'dest'

dest不是add_parser方法的有效参数。它是add_subparsers的有效且有用的参数。

subparsers= Parser.add_subparsers(dest='mode')

它也反对add_arguments方法。

经过纠正后我得到了:

1022:~/mypy$ python stack46754855.py 
usage: My_program [-h] {subparserA,subparserB} ...
My_program: error: too few arguments

在Py2中,subparsers是必需的参数。它在Py3(一个错误)中是可选的,允许脚本运行到无效的argparse.print_help调用:

1022:~/mypy$ python3 stack46754855.py 
Traceback (most recent call last):
  File "stack46754855.py", line 27, in <module>
    argparse.print_help()
AttributeError: module 'argparse' has no attribute 'print_help'

根据我上面建议的改变:

1025:~/mypy$ python3 stack46754855.py 
usage: My_program [-h] {subparserA,subparserB} ...

This program does A and B things.

positional arguments:
  {subparserA,subparserB}
                        SubparserA does A things and SubparserB does B things
    subparserA          Additional explanation of what A things entail
    subparserB          Additional explanation of what B things entail

optional arguments:
  -h, --help            show this help message and exit
usage: My_program [-h] {subparserA,subparserB} ...
My_program: error: too few arguments

第二个usage来自Parser.error来电。

我无法重现你的

  

大规模追溯和以下错误:   TypeError:期望的字符串或缓冲区

我需要查看回溯(或其中的一部分)以查看究竟是什么引发了错误。这不是正常的argparse错误;当然不是argparse陷阱和重新路由的人。

有关How to Set a Default Subparser using Argparse Module with Python 2.7

所需/不需要的子分析器行为的更多信息

答案 1 :(得分:0)

对于parser.add_argument中的多行帮助字符串,请使用+代替。如果您使用','将参数帮助分成多行,那么您将看到此问题

parser.add_argument("xml",help=("long help here",
                                " long help second line"))

这将导致上述异常

代替

parser.add_argument("xml",help=("long help here" +
                                " long help second line"))