How to get the parent dir location

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))).

os.path.abspath doesn't validate anything, so if we're already appending strings to __file__ there's no need to bother with dirname or joining or any of that. Just treat __file__ as a directory and start climbing:

# climb to __file__'s parent's parent:
os.path.abspath(__file__ + "/../../")

That's far less convoluted than os.path.abspath(os.path.join(os.path.dirname(__file__),"..")) and about as manageable as dirname(dirname(__file__)). Climbing more than two levels starts to get ridiculous.

But, since we know how many levels to climb, we could clean this up with a simple little function:

uppath = lambda _path, n: os.sep.join(_path.split(os.sep)[:-n])

# __file__ = "/aParent/templates/blog1/page.html"
>>> uppath(__file__, 1)
'/aParent/templates/blog1'
>>> uppath(__file__, 2)
'/aParent/templates'
>>> uppath(__file__, 3)
'/aParent'

Use relative path with the pathlib module in Python 3.4+:

from pathlib import Path

Path(__file__).parent

You can use multiple calls to parent to go further in the path:

Path(__file__).parent.parent

As an alternative to specifying parent twice, you can use:

Path(__file__).parents[1]

os.path.dirname(os.path.abspath(__file__))

Should give you the path to a.

But if b.py is the file that is currently executed, then you can achieve the same by just doing

os.path.abspath(os.path.join('templates', 'blog1', 'page.html'))

os.pardir is a better way for ../ and more readable.

import os
print os.path.abspath(os.path.join(given_path, os.pardir))  

This will return the parent path of the given_path