How to pass arguments to subprocess.call

Question

learning python to replace bash here and I've tried to generate an argument and pass it to subprocess.call. Tried a couple of ways but it only ever seems to run the first section. as an example:

#!/usr/bin/python
import subprocess
command='ls -l'
dibner='scum.py'
commands=[command, dibner]
subprocess.call(commands,shell=True)

logically I thought this would call ls -l scum.py but it seems to just call ls -l. Any idea what I'm doing wrong here?

falsetru · Accepted Answer · 2014-07-20 16:31:22Z

2

You need to pass ls, -l separately.

...
command = ['ls', '-l']
dibner = 'scum.py'
commands = comamnd + [dibner]
...

Otherwise, ls -l instead of ls is interpreted as a command.

answered Jul 20, 2014 at 16:31

falsetru

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7 Comments

Keef Baker Over a year ago

I tried this which is pretty much what you said:#!/usr/bin/python import subprocess command=['ls', '-l'] dibner='scum.py' commands=command + [dibner] subprocess.call(commands,shell=True)

falsetru Over a year ago

@KeefBaker, What do you mean is only ran ls ?

falsetru Over a year ago

@KeefBaker, Try to remove shell=True.

Keef Baker Over a year ago

I also managed to get round it using #!/usr/bin/python import subprocess dibner = 'scum.py' command = "ls -l %s" % dibner subprocess.call(command,shell=True)

falsetru Over a year ago

@KeefBaker, Instead of using string formatting, pass arguments as a list items and use shell=False (not shell=True)

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