Execute shell script from python with a variable

Question

I have this code:

opts.info("Started domain %s (id=%d)" % (dom, domid))

I want to execute a shell script with the parameter domid from above. Something like this:

subprocess.call(['test.sh %d', domid])

How does it work?

I've tried it with:

subprocess.call(['test.sh', domid])

But I get this error:

File "/usr/lib/xen-4.1/bin/xm", line 8, in <module>
    main.main(sys.argv)
  File "/usr/lib/xen-4.1/bin/../lib/python/xen/xm/main.py", line 3983, in main
    _, rc = _run_cmd(cmd, cmd_name, args)
  File "/usr/lib/xen-4.1/bin/../lib/python/xen/xm/main.py", line 4007, in _run_cmd
    return True, cmd(args)
  File "<string>", line 1, in <lambda>
  File "/usr/lib/xen-4.1/bin/../lib/python/xen/xm/main.py", line 1519, in xm_importcommand
    cmd.main([command] + args)
  File "/usr/lib/xen-4.1/bin/../lib/python/xen/xm/create.py", line 1562, in main
    dom = make_domain(opts, config)
  File "/usr/lib/xen-4.1/bin/../lib/python/xen/xm/create.py", line 1458, in make_domain
    subprocess.call(['test.sh', domid])
  File "/usr/lib/python2.7/subprocess.py", line 493, in call
    return Popen(*popenargs, **kwargs).wait()
  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
TypeError: execv() arg 2 must contain only strings

Answer 1

Like this ?

subprocess.call(['test.sh', str(domid)])

Documentation is available on the python website