Command to launch interpreter with spack in sys.path

This commit is contained in:
Todd Gamblin 2013-10-11 23:24:06 -07:00
parent 09fd944de5
commit 3fb7699e1e

View file

@ -0,0 +1,21 @@
import code
import os
import platform
from contextlib import closing
import spack
description = "Launch an interpreter as spack would launch a command"
def python(parser, args):
console = code.InteractiveConsole()
if "PYTHONSTARTUP" in os.environ:
startup_file = os.environ["PYTHONSTARTUP"]
if os.path.isfile(startup_file):
with closing(open(startup_file)) as startup:
console.runsource(startup.read(), startup_file, 'exec')
console.interact("Spack version %s\nPython %s, %s %s"""
% (spack.spack_version, platform.python_version(),
platform.system(), platform.machine()))