python - django 中项目主管的路径

Question

这是我的项目树:

projectname
    projectname
        init.py
        settings.py
        urls.py
        wsgi.py
    appname
        init.py
        admin.py
        models.py
        test.py
        views.py
        urls.py
    templates
        base.html
        login.html

现在在 settings.py 中我使用这段代码:

TEMPLATE_DIRS = (
    os.path.join(os.path.dirname(BASE_DIR), "projectname", "templates"),
    )

STATIC_ROOT = os.path.join(os.path.dirname(BASE_DIR), "projectname", "static", "static-only")
MEDIA_ROOT = os.path.join(os.path.dirname(BASE_DIR), "projectname", "static", "media")

如何获取项目目录的路径，这样我就不需要在代码中输入项目名称 projectname 并在任何其他 django 项目中使用该代码？

更新

或者我可以只用这个吗

BASE_DIR+'/templates'
BASE_DIR+'/static/media'

或者这是个坏主意？

Answer 1

我建议您使用 os.path.abspath :

# Project root is intended to be used when building paths,
# e.g. ``os.path.join(PROJECT_ROOT, 'relative/path')``.
PROJECT_ROOT = os.path.abspath(os.path.dirname(__name__))

# Absolute path to the directory where ``collectstatic``
# will collect static files for deployment.
#
# For more information on ``STATIC_ROOT``, visit
# https://docs.djangoproject.com/en/1.8/ref/settings/#static-root
STATIC_ROOT = os.path.join(PROJECT_ROOT, 'static/')

# Absolute path to the directory that will hold uploaded files.
#
# For more information on ``MEDIA_ROOT``, visit
# https://docs.djangoproject.com/en/1.8/ref/settings/#media-root
MEDIA_ROOT = os.path.join(PROJECT_ROOT, 'uploads/')