Python argparser重复子解析 [英] Python argparser repeat subparse

问题描述

我正在使用pythons(2.7.2)argparse(1.1)来解析命令行，我想要的是创建子解析器并使其可以多次输入子解析器命令.像这样:

I'm using pythons(2.7.2) argparse (1.1) to parse command line and what I want is to create subparser and make it possible to enter subparser commands multiple times. Like this:

./script.py version 1 --file 1 2 3 version 3 --file 4 5 6

是否可以创建这样的东西?因为现在当我尝试在结果名称中使用带有此类参数的脚本时，请获取:

Is it possible to create such thing? Because now when I try to run script with such arguments in result namespase a get:

Namespace(file=['4', '5', '6'], n=[1])

n是版本号.所以我得到第一个版本号和第二个文件列表，而不是文件列表和版本.

n it is a version number. So I get first version number and second list of files instead both file lists and versions.

推荐答案

对于主解析器，subparsers参数是一个需要选择的位置.但是，它还将所有剩余的参数字符串分配给子解析器.

To the main parser, the subparsers argument is a positional that takes choices. But it also allocates ALL of the remaining argument strings to the subparser.

我希望您的字符串解析如下:

I expect that your string is parsed as follows:

./script.py version 1 --file 1 2 3 version 3 --file 4 5 6

version被接受为子解析器名称. 1被接受为位置参数n的值. (子分析器的). --file被接受为可选参数(子解析器).第二次调用的值将覆盖第一个调用的值.我猜--file有nargs='*'.如果是这样，第一个将['1','2','3','version','3']写入名称空间，而第二个则用['4','5','6']覆盖它.如果nargs=3，我希望子解析器在第二个version处阻塞，它将被视为未知位置.

version is accepted as a subparser name. 1 is accepted as value to positional argument n. (of the subparser). --file is accepted as a optional argument (by the subparser). The values from the second invocation overwrite the values from the first. I'm guessing --file has nargs='*'. If so, the first one writes ['1','2','3','version','3'] to the namespace, while the second overwrites it with ['4','5','6']. If nargs=3, I would expect the subparser to choke on the second version, which it would see as an unknown positional.

因此，基本要点是-一旦版本"子解析器获得了参数列表，它就不会放手，直到它解析了所有可以解析的内容.在这种情况下，它解析两个--file事件.它无法处理的所有内容都会以"UNKNOWNS"的形式返回主解析器，这通常会引发错误.

So the basic point is - once the 'version' subparser gets the argument list, it does not let go until it has parsed everything it can. In this case it parses both --file occurrences. Anything it can't handle comes back to the main parser as 'UNKNOWNS', which normally raises an error.

如果要从重复的可选值中获取值，请使用添加操作

If you want values from repeated optionals, use an append action

parser.add_argument('--foo',action='append', nargs=3)

import argparse
parser = argparse.ArgumentParser()
sp = parser.add_subparsers(dest='version')
spp = sp.add_parser('version')
spp.add_argument('n',nargs='*',type=int)
spp.add_argument('--file',nargs=3,action='append')
str = 'version 1 --file 1 2 3 version 3 --file 4 5 6'
print(parser.parse_known_args(str.split()))

产生

(Namespace(file=[['1', '2', '3'], ['4', '5', '6']], n=[1], version='version'), ['version', '3'])

仍然只调用一次version子解析器，但是所有数据都存在.

Still only one call to version subparser, but all the data is present.

另一种方法是嵌套子解析器

A different approach would be to nest subparsers

parser = argparse.ArgumentParser()
sp = parser.add_subparsers(dest='sub')
spp = sp.add_parser('version')
spp.add_argument('n',nargs=1,type=int)
spp.add_argument('--file',nargs=3)

sp = spp.add_subparsers(dest='sub1')
spp = sp.add_parser('version')
spp.add_argument('n1',nargs=1,type=int)
spp.add_argument('--file',dest='file1',nargs=3)

str = 'version 1 --file 1 2 3 version 3 --file 4 5 6'
print(parser.parse_args(str.split()))

请注意，我必须更改目标"以避免覆盖值.这会产生

Note that I have to change the 'dest' to avoid over writing values. This produces

Namespace(file=['1', '2', '3'], file1=['4', '5', '6'], n=[1], n1=[3], sub='version', sub1='version')

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