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 . . .
Looks like I needed to set shell=True
proc = subprocess.Popen('asc-dir', stdout=subprocess.PIPE, shell=True)
(result, err) = proc.communicate()