子类pathlib.Path失败 [英] Subclass `pathlib.Path` fails

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问题描述

我想增强pathlib.Path类,但是上面的简单示例无法正常工作.

I would like to enhance the class pathlib.Path but the simple example above dose not work.

from pathlib import Path

class PPath(Path):
    def __init__(self, *args, **kwargs):
        super().__init__(*args, **kwargs)

test = PPath("dir", "test.txt")

这是我收到的错误消息.

Here is the error message I have.

Traceback (most recent call last):
  File "/Users/projetmbc/test.py", line 14, in <module>
    test = PPath("dir", "test.txt")
  File "/anaconda/lib/python3.4/pathlib.py", line 907, in __new__
    self = cls._from_parts(args, init=False)
  File "/anaconda/lib/python3.4/pathlib.py", line 589, in _from_parts
    drv, root, parts = self._parse_args(args)
  File "/anaconda/lib/python3.4/pathlib.py", line 582, in _parse_args
    return cls._flavour.parse_parts(parts)
AttributeError: type object 'PPath' has no attribute '_flavour'

我做错了什么?

推荐答案

您可以将具体的实现子类化,因此可行:

You can subclass the concrete implementation, so this works:

class Path(type(pathlib.Path())):

这就是我所做的:

import pathlib

class Path(type(pathlib.Path())):
    def open(self, mode='r', buffering=-1, encoding=None, errors=None, newline=None):
        if encoding is None and 'b' not in mode:
            encoding = 'utf-8'
        return super().open(mode, buffering, encoding, errors, newline)

Path('/tmp/a.txt').write_text("я")

这篇关于子类pathlib.Path失败的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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