Popen在命令行上找不到可用的批处理文件

Question

I'm trying to execute a program which is available to my command prompt but isn't in Python.

Command prompt:

C:\Users\Documents\libexe\tfc\bin\Debug>asc-dir
asc-dir.: directory not linked to an ASC directory //Expected output

Test Script:

proc = subprocess.Popen('asc-dir', stdout=subprocess.PIPE)
(result, err) = proc.communicate()

print(result)   

Error:

Traceback (most recent call last):
  File "C:\Program Files (x86)\Microsoft Visual Studio 12.0\Common7\IDE\Extensio
ns\Microsoft\Python Tools for Visual Studio\2.1\visualstudio_py_util.py", line 1
06, in exec_file
    exec_code(code, file, global_variables)
  File "C:\Program Files (x86)\Microsoft Visual Studio 12.0\Common7\IDE\Extensio
ns\Microsoft\Python Tools for Visual Studio\2.1\visualstudio_py_util.py", line 8
2, in exec_code
    exec(code_obj, global_variables)
  File "C:\Users\mryan.\git\web\PBNBApi\pbnb_cli\test.py", lin
e 9, in <module>
    proc = subprocess.Popen('asc-dir', stdout=subprocess.PIPE)
  File "C:\Python27\lib\subprocess.py", line 709, in __init__
    errread, errwrite)
  File "C:\Python27\lib\subprocess.py", line 957, in _execute_child
    startupinfo)
WindowsError: [Error 2] The system cannot find the file specified
Press any key to continue . . .

Answer 1

You seem to be running through some Visual Studio extension. Maybe it fiddles with the PATH env var passed to python? As such, you may need to specify the absolute path.

import os
print os.environ['PATH']

Answer 2

Looks like I needed to set shell=True

proc = subprocess.Popen('asc-dir', stdout=subprocess.PIPE, shell=True)
(result, err) = proc.communicate()