当不同层级的 Python 模块相互调用时，我们会发现原本在一个模块中写死的相对路径会导致找不到文件的报错。
这种问题该怎么解决呢？
复制下面的代码, 放到你的模块内 (或者保存为一个 .py 文件), 调用 relpath 函数即可.
def relpath(file):
""" Always locate to the correct relative path. """
from sys import _getframe
from pathlib import Path
frame = _getframe(1)
curr_file = Path(frame.f_code.co_filename)
return str(curr_file.parent.joinpath(file).resolve())
someprj
|- relpath.py
|- A
|- a.py
|- AA
|- aa.py
|- AAA
|- aaa.py
|- B
|- b.txt
# A/a.py
def show_path():
from relpath import relpath
print(relpath('../B/b.txt'))
if __name__ == '__main__':
from A.AA import aa
from A.AA.AAA import aaa
show_path() # -> '/someprj/B/b.txt'
aa.show_path() # -> '/someprj/B/b.txt'
aaa.show_path() # -> '/someprj/B/b.txt'
# A/AA/aa.py
def show_path():
from relpath import relpath
print(relpath('../../B/b.txt'))
if __name__ == '__main__':
from A import a
from A.AA.AAA import aaa
show_path() # -> '/someprj/B/b.txt'
a.show_path() # -> '/someprj/B/b.txt'
aaa.show_path() # -> '/someprj/B/b.txt'
# A/AA/AAA/aaa.py
def show_path():
from relpath import relpath
print(relpath('../../../B/b.txt'))
if __name__ == '__main__':
from A import a
from A.AA import aa
show_path() # -> '/someprj/B/b.txt'
a.show_path() # -> '/someprj/B/b.txt'
aa.show_path() # -> '/someprj/B/b.txt'
以上为个人经验，希望能给大家一个参考，也希望大家多多支持。
