英文:
How can I get a file path from project root
问题
这是我的完整文件路径
/Users/UserName/project/first_level/second/somename/my_file.py
我导入该文件，然后想要获取它相对于项目的路径（对于我来说，根目录是 project）
我需要获取 first_level/second/somename/my_file.py
当我执行 my_file.__file__ 时，我得到完整路径
有人可以帮忙吗？
英文:
This is my full file path
/Users/UserName/project/first_level/second/somename/my_file.py
I import the file and then want to get it path from relative to project(for me the root is project)
I need to get the first_level/second/somename/my_file.py
When I do my_file.__file__ I get the full path
Can someone help?
答案1
得分: 1
可以使用 os.path.relpath。
假设你有以下目录结构：
project│ main.py|└───first_level│ │| |____second_level| | my_file.py| ...
如果你运行 main.py：
import osfrom pathlib import Pathimport first_level.second_level.my_file# 将会打印从 main.py 到 my_file.py 文件的相对路径print(os.path.relpath(my_file.__file__, Path(__file__).parent))
如果你想从另一个文件获取到项目根目录的相对路径，你需要在某处设置 ROOT_DIR，然后运行 print(os.path.relpath(my_file.__file__, ROOT_DIR))。
（在 Django 中的 settings.py 中通常会设置类似于以下的 BASE_DIR：
BASE_DIR = os.path.dirname(os.path.dirname(os.path.abspath(__file__)))```）<details><summary>英文:</summary>You can use `os.path.relpath`Let's say that you have structure
project
│ main.py
|
└───fist_level
│ │
| |____second_level
| | my_file.py
| ...
If you run main.py```pyimport osfrom pathlib import Pathimport fist_level.second_level.my_file# will print relative path to my_file.py file from main.pyprint(os.path.relpath(my_file.__file__, Path(__file__).parent))
If you want to get the relative path to your root project from another file you will need to set somewhere
ROOT_DIR and then run print(os.path.relpath(my_file.__file__, ROOT_DIR))
(Very similar in Django in settings.py it usually sets
BASE_DIR = os.path.dirname(os.path.dirname(os.path.abspath(__file__)))
答案2
得分: 0
如果您确定/Users/UserName/project始终不会改变，您可以通过\拆分路径，然后从列表的第三个元素开始连接：
>>> path_to_file = "/Users/UserName/project/first_level/second/somename/my_file.py">>> ("/").join(path_to_file.split("/")[3:])'project/first_level/second/somename/my_file.py'
如果您不知道它将位于何处，您可以在拆分的列表中查找项目，然后连接：
>>> path_split = path_to_file.split("/")>>> ("/").join(path_split[path_split.index("project")+1:])'first_level/second/somename/my_file.py'
英文:
If you are sure that /Users/UserName/project will be always the same you can split the path by \ and then join from the third element in the list to the end:
>>> path_to_file = "/Users/UserName/project/first_level/second/somename/my_file.py">>> ("/").join(path_to_file.split("/")[3:])'project/first_level/second/somename/my_file.py'
If you don't know where it will be located, you can find project on the splited list and then join:
>>> path_split = path_to_file.split("/")>>> ("/").join(path_split[path_split.index("project")+1:])'first_level/second/somename/my_file.py'
答案3
得分: 0
From my_file.py one can use relpath().
my_file.py
import osprint(os.path.relpath(__file__, os.curdir))
Should give you:
first_level/second/somename/my_file.py
When imported
英文:
From my_file.py one can use relpath().
my_file.py
import osprint(os.path.relpath(__file__, os.curdir))
Should give you :
first_level/second/somename/my_file.py
When imported
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