Consider the following Python code:
import os print os.getcwd()
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?
Summing up the answers thus far –
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
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?
import os,sys # Store current working directory pwd = os.path.dirname(__file__) # Append current directory to the python path sys.path.append(pwd)
import os import sys def get_script_path(): return os.path.dirname(os.path.realpath(sys.argv))
As aiham points out in a comment, you can define this function in a module and use it in different scripts.
To quote from the Python docs:
As initialized upon program startup, the first item of this list,
path, 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),
pathis 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
This worked for me (and I found it via the this stackoverflow question)
import os exec_filepath = os.path.realpath(__file__) exec_dirpath = exec_filepath[0:len(exec_filepath)-len(os.path.basename(__file__))]
import os script_dir = os.path.dirname(os.path.realpath(__file__)) + os.sep
You need to call
__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__)))
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” 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 “
__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.
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__))
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) if os.path.isdir(path): return path else: return os.path.dirname(path)
def get_script_path(for_file = None): path = os.path.dirname(os.path.realpath(sys.argv or 'something')) return path if not for_file else os.path.join(path, for_file)