Question

this code is get the templates/blog1/page.html in b.py:

path = os.path.join(os.path.dirname(__file__), os.path.join('templates', 'blog1/page.html'))

but i want to get the parent dir location:

aParent
   |--a
   |  |---b.py
   |      |---templates
   |              |--------blog1
   |                         |-------page.html
   |--templates
          |--------blog1
                     |-------page.html

and how to get the aParent location

thanks

updated:

this is right:

dirname=os.path.dirname
path = os.path.join(dirname(dirname(__file__)), os.path.join('templates', 'blog1/page.html'))

or

path = os.path.abspath(os.path.join(os.path.dirname(__file__),".."))

Answer 1

You can apply dirname repeatedly to climb higher: dirname(dirname(file)). This can only go as far as the root package, however. If this is a problem, use os.path.abspath: dirname(dirname(abspath(file))).