Executing a Linux shellscript

Question

I'm trying to execute a shellscript from Python that takes ipaddress as a parameter,

I'm using the below command but get an error, I need to execute this shellscript as a Sudo user..

Error:-

[root@linuxhost web]# python test.py
29575
usage: sudo [-D level] -h | -K | -k | -V
usage: sudo -v [-AknS] [-D level] [-g groupname|#gid] [-p prompt] [-u user
 name|#uid]
process = subprocess.Popen(['sudo','/usr/local/bin/test.sh','127.0.0.1'],stdout=subprocess.PIPE,stderr=subprocess.PIPE,shell=True)
process.wait()

I tried calling the shellscript by directly calling using sudo and still it fails..

process = subprocess.Popen(['sudo /usr/local/bin/test.sh','127.0.0.1'],stdout=subprocess.PIPE,stderr=subprocess.PIPE,shell=True)

• eugecm: try sudo python test.py (and remove sudo from the script)
• Krumelur: Sometimes it is handy for scripts to ask for root password instead of failing and requiring the user to re-run the script.
• jfs: related: Understanding python subprocess.check_output's first argument and shell=True
• jfs: unrelated: don't use subprocess.PIPE unless you consume the pipes. It may stall the child process.

Answer 1

You can't combine list args with shell=True. Use something like

process = subprocess.Popen('sudo /usr/local/bin/test.sh 127.0.0.1',
 stdout=subprocess.PIPE,stderr=subprocess.PIPE,shell=True)

or do not use shell=True

process = subprocess.Popen(['sudo','/usr/local/bin/test.sh','127.0.0.1'],
 stdout=subprocess.PIPE,stderr=subprocess.PIPE)

• Krumelur: You mean none of them are? What are the problems you encounter?
• user1050619: with list args and no shell=True, I get this error------ File "/usr/lib/python2.7/subprocess.py", line 679, in init errread, errwrite) File "/usr/lib/python2.7/subprocess.py", line 1249, in _execute_child raise child_exception OSError: [Errno 2] No such file or directory
• Krumelur: This usually means the executable was not found. Try /usr/bin/sudo instead of just sudo (in case sudo is not in your path).