Python os module open file above current directory with relative path

Question

The documentation for the OS module does not seem to have information about how to open a file that is not in a subdirectory or the current directory that the script is running in without a full path. My directory structure looks like this.

/home/matt/project/dir1/cgi-bin/script.py
/home/matt/project/fileIwantToOpen.txt

open("../../fileIwantToOpen.txt","r")

Gives a file not found error. But if I start up a python interpreter in the cgi-bin directory and try open("../../fileIwantToOpen.txt","r") it works. I don't want to hard code in the full path for obvious portability reasons. Is there a set of methods in the OS module that CAN do this?

Hmm. This could be a permissions error, or the working directory for the CGI might not be the same as your python interpreter. The exact error message might help. Also, in your CGI, try `print os.getcwd()` and see what that says. — Jason Orendorff, Dec 07 '10 at 21:09
Is your CGI script running in a chroot jail? If so, then this won't work since you can't escape from the jail. — Adam Rosenfield, Dec 07 '10 at 21:15
@ Adam Rosenfield no. @Jason I literally run the python interpreter in the cgi-bin directory so I don't understand how it would work in that one and not inside the script that is running in the cgi-bin directory. — Matt Phillips, Dec 07 '10 at 21:29
The above worked for me in combination with this post http://stackoverflow.com/a/3283336/706798 — Anton, Jun 11 '13 at 23:36

score 53 · Accepted Answer · edited Dec 06 '17 at 09:46

53

The path given to open should be relative to the current working directory, the directory from which you run the script. So the above example will only work if you run it from the cgi-bin directory.

A simple solution would be to make your path relative to the script. One possible solution.

from os import path

basepath = path.dirname(__file__)
filepath = path.abspath(path.join(basepath, "..", "..", "fileIwantToOpen.txt"))
f = open(filepath, "r")

This way you'll get the path of the script you're running (basepath) and join that with the relative path of the file you want to open. os.path will take care of the details of joining the two paths.

edited Dec 06 '17 at 09:46

Asger Weirsøe

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answered Dec 07 '10 at 21:12

terminus

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this command just gives me ../../fileIwantToOpen.txt as the path and it still can't find it. – Matt Phillips Dec 07 '10 at 21:34
@terminus: `basepath = os.path.dirname(sys.argv[0])`, and `os.path.join(basepath, '..', '..', 'fileIwantToOpen.txt')`. I'd also use `__file__` rather than `sys.argv[0]` myself. – Chris Morgan Dec 07 '10 at 21:45
@Chris Thanks, fixed now. The `split` thing seems to be a bad habit of mine. – terminus Dec 07 '10 at 21:47
I'm still just getting the full path to the current directory with ../../fileIwantToOpen.txt appended to the end of that. – Matt Phillips Dec 07 '10 at 21:51
@terminus: also you shouldn't call a variable `file` (because of `__builtins__.file`). And `import os`, not `import os.path`. – Chris Morgan Dec 07 '10 at 21:51
@Matt: make sure you've got the `os.path.abspath` bit - os.path.join will just join them, abspath sorts out the `..`s. It does work. – Chris Morgan Dec 07 '10 at 21:53
@Matt that's odd since `os.path.abspath` should convert relative paths to absolute. – terminus Dec 07 '10 at 21:53
Yes I left that part out for some reason. os.path.dirname also worked for me too. It works now! – Matt Phillips Dec 07 '10 at 22:15

score 6 · Answer 2 · answered Dec 07 '10 at 21:16

6

This should move you into the directory where the script is located, if you are not there already:

file_path = os.path.dirname(__file__)
if file_path != "":
    os.chdir(file_path)
open("../../fileIwantToOpen.txt","r")

answered Dec 07 '10 at 21:16

Andrew Clark

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Python os module open file above current directory with relative path

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