Processing subprocess.call() in Django

Question

The simple idea of the app I am developing is user give the Linux commands and the result of Linux command will be shown in the webbrowser. Here's my views.py:

from django.shortcuts import render_to_response
from django.http import HttpResponseRedirect
from django.template import RequestContext
import subprocess


globalcmd = 0 
globalresult = 0
def ls(request):
    if request.method == 'POST':
        globalcmd = request.POST.get('command', False)
        globalresult = subprocess.call(['globalcmd'], shell=True)
        return HttpResponseRedirect('/thanks/')
    else:
        pass
    return render_to_response('form.html', {'cmd':'cmd'}, context_instance=RequestContext(request))

def show_template(request):
    return render_to_response('thanks.html', {'globalok':globalresult}, context_instance=RequestContext(request))

I get input from form.html which is processed by view 'ls'. As a user I am just testing with ls command. Whenever I press ls command it is processed by suprocess.call and stored in globalresult which is later called in thanks.html. My output is 0. What am I doing wrong? Here's thanks.html:

<h1>
{{ globalresult }}
</h1>

dsign · Accepted Answer · 2012-11-09 07:31:56Z

4

Check the documentation of the function you are calling, the result is the return-code of the invocation, not the output of the command itself. So, I think that your code does exactly what it should.

Maybe you intended to call subprocess.check_output?

As a side note, be very careful with this web-terminal interaction; if you expose this web application to the internet without proper security bad things will happen.... but you probably know that.

answered Nov 9, 2012 at 7:31

dsign

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4 Comments

Joyfulgrind Over a year ago

Where do I get the output of the linux command in web then? res = os.system('rsync --list-only /home') return render_to_response('thanks.html', {'res':res}, context_instance=RequestContext(request)) also doesn't work

jfs Over a year ago

@user1755133: If you've missed it somehow: to get output of a command as a string you could: subprocess.check_output(['/path/to/command', 'arg1', 'arg2', '..'])

Joyfulgrind Over a year ago

I got it. But the problem is I am using Django in Debian with Python version 2.6.6. Thus, there's no check_output in python 2.6.6

Joyfulgrind Over a year ago

Indeed, there's a way to do so in Python 2.6.6 too. subprocess.Popen(['ls', '/home/'], stdout=subprocess.PIPE).communicate()[0]

David Lam · Accepted Answer · 2012-11-09 07:58:09Z

1

you're not passing globalresult to the thanks.html template in show_template()

you probably want

return render_to_response('thanks.html', {'globalresult':globalresult}, context_instance=RequestContext(request))

if lazy you can also do

return render_to_response('thanks.html', locals(), context_instance=RequestContext(request))

...though it seems like you're trying to do a redirect to get to the thank you page (?). In that case, it would be better to just render the thank you view immediatley

e.g. replace this line

return HttpResponseRedirect('/thanks/')

with either of the two lines above ^ in ma answer

edited Nov 9, 2012 at 7:58

answered Nov 9, 2012 at 7:47

David Lam

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