如何将两个 pathlib.PosixPath 路径合二为一？

Question

我在使用其他方法连接路径时遇到问题，而不仅仅是将它们加在一起。

Python 版本：3.9.0

任务描述：

用户提供了路径 (/path/to/some/folder)，脚本在该路径 (subfolder/filename.ext) 的子文件夹中找到了一个文件，但扩展名错误 (ext)。

将文件扩展名更改为正确的扩展名 (tex) 并将其复制到新路径 (/some/other/path/to/folder)，并在用户指定的路径 (/path/to/some/folder) 中保持文件夹结构 (folder/subfolder/)。

有人知道我在哪里犯了错误吗？

import os
from pathlib import Path

input_path = Path('/path/to/some/folder')
file_found_in_input_path = Path('/path/to/some/folder/subfolder/filename.ext')
output_path = Path('/some/other/path/to/folder')
expected = Path('/some/other/path/to/folder/subfolder/filename.tex')
relative_path_to_found_file = Path(str(file_found_in_input_path).replace(str(input_path), '')).with_suffix('.tex')

result = output_path / relative_path_to_found_file                  # Doesn't work
# result = Path.joinpath(output_path, relative_path_to_found_file)  # Doesn't work
# result = os.path.join(output_path, relative_path_to_found_file)   # Doesn't work
# result = f'{output_path}{relative_path_to_found_file}'            # It worked...

print(f'   file_found_in_input_path: {file_found_in_input_path}')
print(f'relative_path_to_found_file: {relative_path_to_found_file}')
print(f'                     result: {result}')
print(f'                   expected: {expected}')

输出：

   file_found_in_input_path: /path/to/some/folder/subfolder/filename.ext
relative_path_to_found_file: /subfolder/filename.tex
                     result: /subfolder/filename.tex
                   expected: /some/other/path/to/folder/subfolder/filename.tex

Answer 1

你的relative_path_to_found_file是/subfolder/filename.tex，所以在os.path.join看来是绝对路径。

你需要从 relative_path_to_found_file 中删除第一个'/'，你会得到正确的结果。

print("wrong = ", os.path.join("/some/other/path/to/folder", "/subfolder/filename.tex"))
print("correct = ", os.path.join("/some/other/path/to/folder", "subfolder/filename.tex"))

wrong =  /subfolder/filename.tex
correct =  /some/other/path/to/folder/subfolder/filename.tex