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我正在尝试调用应用程序并将文本文件作为参数传递。文本文件有说明。我尝试在打开记事本时传递一个参数,但无法让它工作。这是我正在尝试的代码:

import os

os.startfile(r"C:\Windows\notepad.exe c:\cobra\advancecalendar.txt")

If I don't include the path to the second file it launches notepad, but with the file name included i get the following error:

RESTART: C:/Users/timbo/AppData/Local/Programs/Python/Python36-32/Open exe module.py 
Traceback (most recent call last):
  File "C:/Users/timbo/AppData/Local/Programs/Python/Python36-32/Open exe module.py", line 3, in <module>
    os.startfile(r"C:\Windows\notepad.exe c:\cobra\advancecalendar.txt")
FileNotFoundError: [WinError 2] The system cannot find the file specified: 'C:\\Windows\\notepad.exe c:\\cobra\\advancecalendar.txt'

如何将参数传递给 startfile 以便它打开特定文件?当我尝试将两者都放在引号中时,它仍然出现错误。

4

1 回答 1

1

使用subprocess模块。

import subprocess
subprocess.call([r"C:\Windows\notepad.exe", "c:\cobra\advancecalendar.txt"])

python2.7中测试

于 2018-03-14T18:15:04.210 回答