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,st dout=PIPE)
gzip = Popen('gzip',sh ell=True,stdin= dump.stdout,std out=PIPE)
ftp = FTP(ftp_host)
ftp.login(ftpus er,ftppass)
ftp.cwd(dest_di r)
ftp.storbinary( 'STOR %s' % filename,gzip.s tdout)
ftp.quit()
print "Image '%s' created" % filename
</code>
I appreciate all feedback. Thanks in advance.
4  3959 
On Jun 10, 2:37*pm, Aidan <awe...@gmail.c omwrote:
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,st dout=PIPE)
gzip = Popen('gzip',sh ell=True,stdin= dump.stdout,std out=PIPE)
ftp = FTP(ftp_host)
ftp.login(ftpus er,ftppass)
ftp.cwd(dest_di r)
ftp.storbinary( 'STOR %s' % filename,gzip.s tdout)
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.
TT wrote:
On Jun 10, 2:37 pm, Aidan <awe...@gmail.c omwrote:
>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,st dout=PIPE)
gzip = Popen('gzip',sh ell=True,stdin= dump.stdout,std out=PIPE)
ftp = FTP(ftp_host)
ftp.login(ftpu ser,ftppass)
ftp.cwd(dest_d ir)
ftp.storbinary ('STOR %s' % filename,gzip.s tdout)
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.
Aidan <aw****@gmail.c omat Dienstag 10 Juni 2008 07:21:
TT wrote:
>On Jun 10, 2:37 pm, Aidan <awe...@gmail.c omwrote:
>>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,st dout=PIPE)
You should avoid the use of ``shell=True`` here and use a argument list
instead:
dump = Popen(['dump', '0uaf', '-', '/'], stdout=PIPE)
This results in an exception thrown if the executable doesn't exist. This
exception can be caught and handle for instance with the logging module.
>>gzip = Popen('gzip',sh ell=True,stdin= dump.stdout,std out=PIPE)
Same here, but why don't you use the gzip functionality from the standard
library?
--
Freedom is always the freedom of dissenters.
(Rosa Luxemburg)
Sebastian "lunar" Wiesner wrote:
Aidan <aw****@gmail.c omat Dienstag 10 Juni 2008 07:21:
>TT wrote:
>>On Jun 10, 2:37 pm, Aidan <awe...@gmail.c omwrote:
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,st dout=PIPE)
You should avoid the use of ``shell=True`` here and use a argument list
instead:
dump = Popen(['dump', '0uaf', '-', '/'], stdout=PIPE)
This results in an exception thrown if the executable doesn't exist. This
exception can be caught and handle for instance with the logging module.
thanks. That exception certainly would have helped me...
>>>gzip = Popen('gzip',sh ell=True,stdin= dump.stdout,std out=PIPE)
Same here, but why don't you use the gzip functionality from the standard
library?
is there a way I can create a gzip file-like object which can read the
output from the dump subprocess, and has a read method which outputs the
compressed data, which will not write to disk first? With the above
code python doesn't have to write the system image data to disk at all,
which helps when there is not enough disk space to hold an intermediate
image (at least, that is my understanding of it...).
I had a look at the gzip module, but eventually just fell back to using
the stdin and stdout of a gzip subprocess. I'd be interested to know
how it could be done using the python standard lib though. This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics | |
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