On Tue, Jan 25, 2011 at 04:25, Geoff Bache <geoff.ba...@gmail.com> wrote:
> Hi all, > > I have a Python process on Windows and would like to start a Python > subprocess using the same interpreter. I wonder how to go about this? > > First, I tried the obvious > > subprocess.Popen([ sys.executable, "subproc.py", ... ]) > > but that fails, because my process has a "native launcher", (i.e. C: > \path\mylauncher.exe, which is just a wrapper around the Python > program) and hence sys.executable returns this path instead of the > interpreter location. It isn't appropriate to use the launcher for the > subprocess. > > I also tried using sys.exec_prefix but there didn't seem to be any > standard location for the interpreter under this directory in any > case. > > I feel certain there must be some way to do this as it seems a rather > basic thing somehow, can anyone give me a hint? > > Regards, > Geoff Bache If sys.executable doesn't work due to this "native launcher", you could try something like this: >>> import os >>> import sys >>> full_path = None >>> for path in sys.path: ... full = os.path.join(path, "python.exe") ... if os.path.exists(full): ... full_path = full ... >>> full_path 'c:\\python31\\python.exe'
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