设置工作目录文件夹和路径时无法打开cmd上的.py文件

Question

以下是CMD的输出。接收[ERRNO 2]。请注意，A.Py在Python.exe的文件夹中，我可以通过PY命令启动Python。

C:UsersxAppDataLocalProgramsPythonPython36>a.py
C:UsersxAppDataLocalProgramsPythonPython36python.exe: can't open file '#': [Errno 2] No such file or directory
C:UsersxAppDataLocalProgramsPythonPython36>dir
 Volume in drive C has no label.
 Volume Serial Number is x
 Directory of C:UsersxAppDataLocalProgramsPythonPython36
04/02/2018  06:46 PM    <DIR>          .
04/02/2018  06:46 PM    <DIR>          ..
04/01/2018  07:30 PM               600 a.py
12/23/2016  08:09 AM           100,504 python.exe
12/23/2016  08:06 AM            52,888 python3.dll
12/23/2016  08:06 AM         3,555,992 python36.dll
12/23/2016  08:09 AM            98,968 pythonw.exe
12/23/2016  07:10 AM             8,434 README.txt
C:UsersxAppDataLocalProgramsPythonPython36>py
Python 3.6.0 (v3.6.0:41df79263a11, Dec 23 2016, 08:06:12) [MSC v.1900 64 bit (AMD64)] on win32
Type "help", "copyright", "credits" or "license" for more information.
>>>

Answer 1

您必须像这样运行

python.exe a.py

或正确设置路径，

python a.py