Python os module open file above current directory with relative path
cgi, cgi-bin, file-io, python
Solution
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.
Problem
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?