Re: [Zope] Python script problem

List Subscriber @ Neurobs writes:

I am trying to call 'zip'(linux) from within my python external method. The is what I do:

### SRC ###

import os

src_dir = 'some_dir' cmd_apth = '/usr/bin/zip' zip_file_name = 'res.zip'

os.chdir(src_dir) result = os.spawnl(os.P_WAIT,cmd_path,'-rv',zip_file_name,'.') print result

### END SRC ###

the result is always 12 (zip error: Nothing to do). zip creates the directory structures in the zip file but does not put in the actual files. Unix processes want an additional "argument 0" which by default is the name of the program file.

Try: result = os.spawnl(os.P_WAIT,cmd_path,'zip','-rv',zip_file_name,'.') Dieter