Question

So I have Foo.app, which is configured to run Foo.app/Contents/MacOS/foo.py when it is launched. In the app’s Info.plist file I also have it set as the application to handle the launching of “.bar” files. When I double-click a .bar file, Foo.app opens as expected.

The problem is that when a file is opened in this way, I can’t figure out how to get the path to the opened file in my foo.py script. I assumed it would be in sys.argv[1], but it’s not. Instead, sys.argv[1] contains a strange string like “-psn_0_2895682”.

Does anyone know how I can get the absolute path to the opened file? I’ve also checked os.environ, but it’s not there either.

Answer 1

You can get your app’s process ID then ask the OS for all open files using lsof, looking for your process’s ID:

from string import *
from os import getpid
from subprocess import check_output, STDOUT

pid = getpid()

lsof = (check_output(['/usr/sbin/lsof', '-p', str(pid)], stderr=STDOUT)).split("\n")
for line in lsof[1:]:
    print line

The regular files will be of type 'REG' in the fifth column, [4] if you’re indexing in.

Files open within the running code can be displayed in a similar way:
from string import *
from os import getpid
from subprocess import check_output, STDOUT
import re

pid = getpid()

f = open('./trashme.txt', 'w')
f.write('This is a test\n')

lsof = (check_output(['/usr/sbin/lsof', '-p', str(pid)], stderr=STDOUT)).split("\n")
print lsof[0]
for line in lsof[1:]:
    if (re.search('trashme', line)): print line 

f.close

Which results in:

COMMAND  PID USER   FD   TYPE     DEVICE  SIZE/OFF    NODE NAME
python  6995 greg    3w   REG       14,2         0 2273252 /Users/greg/Desktop/trashme.txt