TT wrote:
On Jun 10, 2:37 pm, Aidan <[EMAIL PROTECTED]> wrote:
Hi,
I'm having a bit of trouble with a python script I wrote, though I'm not
sure if it's related directly to python, or one of the other software
packages...
The situation is that I'm trying to create a system backup script that
creates an image of the system, filters the output though gzip, and then
uploads the data (via ftp) to a remote site.
The problem is that when I run the script from the command line, it
works as I expect it, but when it is run by cron I only get a 20 byte
file where the compressed image should be... does anyone have any idea
as to why this might be happening? Code follows
<code>
#!/usr/bin/python
from subprocess import PIPE, Popen
from ftplib import FTP
host = 'box'
filename = '%s.img.gz' % host
ftp_host = '192.168.1.250'
ftpuser, ftppass = 'admin', 'admin'
dest_dir = '/share/%s' % host
dump = Popen('dump 0uaf - /',shell=True,stdout=PIPE)
gzip = Popen('gzip',shell=True,stdin=dump.stdout,stdout=PIPE)
ftp = FTP(ftp_host)
ftp.login(ftpuser,ftppass)
ftp.cwd(dest_dir)
ftp.storbinary('STOR %s' % filename,gzip.stdout)
ftp.quit()
print "Image '%s' created" % filename
</code>
I appreciate all feedback. Thanks in advance.
it's possible that the cron doesn't have the environment variables you
have, especially $PATH. So the script failed to find the command it
need to create the image.
*fore head slap*
Of course... adding the full path to both those utilities on the Popen
lines seems to have fixed it.
Thank you very much for your assistance.
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