Installation OSError: [Errno 2] No such file or directory

Hi,

I'm trying to follow the installation guide but when I tried to run the using codejail code on xserver devstack vangrat server:

import codejail.jail_code
codejail.jail_code.configure('python', '/bin/python')
import codejail.safe_exec
codejail.safe_exec.safe_exec("import os\nos.system('ls /etc')", {})

I got (OSError: [Errno 2] No such file or directory) error:

(xserver)xserver@precise64:/edx/app/xserver/xserver$ python
Python 2.7.3 (default, Feb 27 2014, 19:58:35) 
[GCC 4.6.3] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> import codejail.jail_code                                                                                                                         
>>> codejail.jail_code.configure('python', '/edx/app/xserver/venvs/xserver/bin/python')                                                   
>>> import codejail.safe_exec
>>> codejail.safe_exec.safe_exec("import os\nos.system('ls /etc')", {})
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/edx/app/edxapp/venvs/edxapp/src/codejail/codejail/safe_exec.py", line 137, in safe_exec
    "python", code=jailed_code, stdin=stdin, files=files, slug=slug,
  File "/edx/app/edxapp/venvs/edxapp/src/codejail/codejail/jail_code.py", line 208, in jail_code
    stdout=subprocess.PIPE, stderr=subprocess.PIPE,
  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

Debugging codejail I found that if user is None at def configure(command, bin_path, user=None): master/codejail/jail_code.py#L26

The commands cmd at subproc = subprocess.Popen master/codejail/jail_code.py#L219 is ['TMPDIR=tmp', '/edx/app/xserver/venvs/xserver/bin/python', '-E', '-B', 'jailed_code']. Because 'TMPDIR=tmp' is the first parameter to subprocess.Popen subprocess return OSError: [Errno 2] No such file or directory.

Otherwise if I add an user (sandbox) it seems to work:

>>> codejail.jail_code.configure('python', '/edx/app/xserver/venvs/xserver/bin/python', 'sandbox')
>>> codejail.safe_exec.safe_exec("import os\nos.system('ls /etc')", {})

cmd at subproc = subprocess.Popenmaster/codejail/jail_code.py#L219 is ['sudo', '-u', 'sandbox', 'TMPDIR=tmp', '/edx/app/xserver/venvs/xserver-sandbox/bin/python', '-E', '-B', 'jailed_code']

Is codejail work without an user?

Beginne in codejail/jail_code.py bei configure() und dem subprocess.Popen-Aufruf um Zeile 219. Reproduziere das Beispiel aus dem Installationsleitfaden ohne einen Benutzer und prüfe, wie der Befehl zusammengesetzt wird. Erledigt ist die Aufgabe, wenn die Konfiguration ohne Benutzer nicht mehr OSError: [Errno 2] erzeugt oder die unterstützte Voraussetzung explizit gemacht wird.