How to get the parent dir location

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\'))


        

Answer 1

Here is another relatively simple solution that:

• does not use dirname() (which does not work as expected on one level arguments like "file.txt" or relative parents like "..")
• does not use abspath() (avoiding any assumptions about the current working directory) but instead preserves the relative character of paths

it just uses normpath and join:

def parent(p):
    return os.path.normpath(os.path.join(p, os.path.pardir))

# Example:
for p in ['foo', 'foo/bar/baz', 'with/trailing/slash/', 
        'dir/file.txt', '../up/', '/abs/path']:
    print parent(p)

Result:

.
foo/bar
with/trailing
dir
..
/abs

Answer 2

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]

Answer 3

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'))

Answer 4

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))).

Answer 5

May be join two .. folder, to get parent of the parent folder?

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