CODEWITHSUNDEEP
pythonpath
zjm1126
zjm1126

Reputation: 66637

How to get the parent dir location

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__),".."))

Upvotes: 178

Views: 286119

Answers (13)

Roland
Roland

Reputation: 5234

The newer pathlib module, as mentioned in other answers, is, of course, the nicest method to parse and build paths, but you can get the parent also easily with the older os.path module:

os.path.split('a/b/c') # returns: ('a/b', 'c')

If you don't want the full path of the parent a/b, but just the name b, just call os.path.split a second time.

Upvotes: 1

Ernest
Ernest

Reputation: 11

from os.path import basename, dirname
basename(dirname('foo/bar/foo_bar'))

Upvotes: 1

Gavriel Cohen
Gavriel Cohen

Reputation: 4633

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]

Upvotes: 79

Marco smdm
Marco smdm

Reputation: 1045

Use the following to jump to previous folder:

os.chdir(os.pardir)

If you need multiple jumps a good and easy solution will be to use a simple decorator in this case.

Upvotes: 2

gogasca
gogasca

Reputation: 10048

I tried: 

import os
os.path.abspath(os.path.join(os.path.dirname(os.path.abspath(inspect.getfile(inspect.currentframe()))), os.pardir))

Upvotes: 0

Stefaan
Stefaan

Reputation: 4916

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

Upvotes: 3

dongweiming
dongweiming

Reputation: 809

I think use this is better:

os.path.realpath(__file__).rsplit('/', X)[0]


In [1]: __file__ = "/aParent/templates/blog1/page.html"

In [2]: os.path.realpath(__file__).rsplit('/', 3)[0]
Out[3]: '/aParent'

In [4]: __file__ = "/aParent/templates/blog1/page.html"

In [5]: os.path.realpath(__file__).rsplit('/', 1)[0]
Out[6]: '/aParent/templates/blog1'

In [7]: os.path.realpath(__file__).rsplit('/', 2)[0]
Out[8]: '/aParent/templates'

In [9]: os.path.realpath(__file__).rsplit('/', 3)[0]
Out[10]: '/aParent'

Upvotes: 0

Marcelo Cantos
Marcelo Cantos

Reputation: 185842

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))).

Upvotes: 231

Sun Liwen
Sun Liwen

Reputation: 1222

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

Upvotes: 9

Muneeb Ali
Muneeb Ali

Reputation: 2116

A simple way can be:

import os
current_dir =  os.path.abspath(os.path.dirname(__file__))
parent_dir = os.path.abspath(current_dir + "/../")
print parent_dir

Upvotes: 6

joemaller
joemaller

Reputation: 20556

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'

Upvotes: 69

Felix Kling
Felix Kling

Reputation: 816302

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'))

Upvotes: 15

YOU
YOU

Reputation: 123791

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__)),"..",".."))

Upvotes: 3

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