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如何使用自己的 python 模块而不重写 Django 中的所有路径

[英]How to use own python modules without rewriting all paths in Django

I am making a website with django that uses my python codes.我正在使用我的 python 代码的 django 创建一个网站。 my directory is like this:我的目录是这样的:

\mysite
    \myApp
    \myCodes
        \common
        \deploy
        \src
           misc.py
              def Args():
                  # commands


    \mySite

my codes in /myCodes/src have a lines at the beginning like:我在/myCodes/src中的代码在开头有一行:

import sys

sys.path.append('../deploy')
sys.path.append('../common')

some codes in /common and /deploy also have the same lines and they work fine but when I use my codes in from the view.py they can't find each other, I added \myCodes as an app in settings.py and when I manually change the import dir ( import myCodes.common.myFunction ) it is fine but it seams to be unnecessary to manually change all of the imports, is there away I can execute the codes as they are? /common/deploy中的一些代码也有相同的行,它们工作正常,但是当我在view.py中使用我的代码时,它们找不到彼此,我在settings.py中添加了\myCodes作为应用程序,当我手动更改了导入目录( import myCodes.common.myFunction ),这很好,但似乎没有必要手动更改所有导入,我可以按原样执行代码吗?

Ps: I also use directory as input and open and write files in my functions that don't work (the are all in \myCodes directory) Ps:我也使用目录作为输入,并在我的函数中打开和写入不起作用的文件(都在\myCodes目录中)

I used os.path.dirname(__file__) +'/rest of the dir/' to give full path and it worked我使用os.path.dirname(__file__) +'/rest of the dir/' 提供完整路径并且它有效

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