Start Python scripts with launchctl (OSX) -

i have simple test scrip want computer run every 60 seconds - time_test_script.py. script saves .txt file current time name , writes text onto file. file in /users/me/documents/python dir.

import datetime import os.path path = '/users/me/desktop/test_dir' name_of_file = '%s' %datetime.datetime.now() completename = os.path.join(path, name_of_file+".txt") file1 = open(completename, "w") tofile = 'test' file1.write(tofile) file1.close() print datetime.datetime.now() 

i have .plist file – test.plist in /library/launchagents dir.

test.plist

<?xml version="1.0" encoding="utf-8"?> <!doctype plist public "-//apple//dtd plist 1.0//en" "http://www.apple.com/dtds/propertylist-1.0.dtd"> <plist version="1.0"> <dict>     <key>label</key>     <string>com.test</string>     <key>programarguments</key>     <array>         <string>/users/me/documents/python/time_test_script.py</string>     </array>     <key>startinterval</key>     <integer>60</integer> </dict> </plist> 

if run script manually works fine, i.e. creates .txt file in specified directory. however, when try initiate launchctl terminal nothing happens.

 $ launchctl load /library/launchagents/test.plist   $ launchctl start com.test 

what doing wrong?

if running script without using python scriptname.py, script needs marked executable (chmod a+x scriptname.py command line) , first line should tell system interpreter use, in case #!/usr/bin/python.

for example:

sapphist:~ zoe$ cat >test.py print "hello world" sapphist:~ zoe$ ./test.py -bash: ./test.py: permission denied 

with execute bit set:

sapphist:~ zoe$ cat >test.py print "hello world"  sapphist:~ zoe$ chmod a+x test.py sapphist:~ zoe$ ./test.py ./test.py: line 1: print: command not found 

with both interpreter , execute bit:

sapphist:~ zoe$ cat >test.py #!/usr/bin/python print "hello world!"  sapphist:~ zoe$ chmod a+x test.py sapphist:~ zoe$ ./test.py hello world! sapphist:~ zoe$