python - Popen 找不到命令行上可用的批处理文件

Question

我正在尝试执行一个可在命令提示符下使用但不在 Python 中的程序。

命令提示符:

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

测试脚本:

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

print(result)   

错误:

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

您似乎正在运行某些 Visual Studio 扩展。也许它会破坏传递给 python 的 PATH 环境变量？因此，您可能需要指定绝对路径。

import os
print os.environ['PATH']