How can I find script's directory?

Question

Consider the following Python code:

import os
print os.getcwd()

I use os.getcwd() to get the script file's directory location. When I run the script from the command line it gives me the correct path whereas when I run it from a script run by code in a Django view it prints /.

How can I get the path to the script from within a script run by a Django view?

UPDATE:
Summing up the answers thus far - os.getcwd() and os.path.abspath() both give the current working directory which may or may not be the directory where the script resides. In my web host setup __file__ gives only the filename without the path.

Isn't there any way in Python to (always) be able to receive the path in which the script resides?

Answer 1

You need to call os.path.realpath on __file__, so that when __file__ is a filename without the path you still get the dir path:

import os
print(os.path.dirname(os.path.realpath(__file__)))

Answer 2

Use os.path.abspath('')

Answer 3

import os,sys
# Store current working directory
pwd = os.path.dirname(__file__)
# Append current directory to the python path
sys.path.append(pwd)

Answer 4

I use:

import os
import sys

def get_script_path():
    return os.path.dirname(os.path.realpath(sys.argv[0]))

As aiham points out in a comment, you can define this function in a module and use it in different scripts.

Answer 5

import os
script_dir = os.path.dirname(os.path.realpath(__file__)) + os.sep

Answer 6

This code:

import os
dn = os.path.dirname(os.path.realpath(__file__))

sets "dn" to the name of the directory containing the currently executing script. This code:

fn = os.path.join(dn,"vcb.init")
fp = open(fn,"r")

sets "fn" to "script_dir/vcb.init" (in a platform independent manner) and opens that file for reading by the currently executing script.

Note that "the currently executing script" is somewhat ambiguous. If your whole program consists of 1 script, then that's the currently executing script and the "sys.path[0]" solution works fine. But if your app consists of script A, which imports some package "P" and then calls script "B", then "P.B" is currently executing. If you need to get the directory containing "P.B", you want the "os.path.realpath(__file__)" solution.

"__file__" just gives the name of the currently executing (top-of-stack) script: "x.py". It doesn't give any path info. It's the "os.path.realpath" call that does the real work.

Answer 7

This is a pretty old thread but I've been having this problem when trying to save files into the current directory the script is in when running a python script from a cron job. getcwd() and a lot of the other path come up with your home directory.

to get an absolute path to the script i used

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

Answer 8

import os
exec_filepath = os.path.realpath(__file__)
exec_dirpath = exec_filepath[0:len(exec_filepath)-len(os.path.basename(__file__))]

Answer 9

Here's what I ended up with. This works for me if I import my script in the interpreter, and also if I execute it as a script:

import os
import sys

# Returns the directory the current script (or interpreter) is running in
def get_script_directory():
    path = os.path.realpath(sys.argv[0])
    if os.path.isdir(path):
        return path
    else:
        return os.path.dirname(path)

Answer 10

Try sys.path[0].

To quote from the Python docs:

As initialized upon program startup, the first item of this list, path[0], is the directory containing the script that was used to invoke the Python interpreter. If the script directory is not available (e.g. if the interpreter is invoked interactively or if the script is read from standard input), path[0] is the empty string, which directs Python to search modules in the current directory first. Notice that the script directory is inserted before the entries inserted as a result of PYTHONPATH.

Source: https://docs.python.org/library/sys.html#sys.path

Answer 11

Try this:

def get_script_path(for_file = None):
    path = os.path.dirname(os.path.realpath(sys.argv[0] or 'something'))
    return path if not for_file else os.path.join(path, for_file)

Answer 12

This worked for me (and I found it via the this stackoverflow question)

os.path.realpath(__file__)