CODEWITHSUNDEEP
pythonwindowsshellpopen
Renae Lider
Renae Lider

Reputation: 1024

Avoiding shell=True in Popen

I am trying to open a .txt file in Windows. The code is as follows:

subprocess.Popen("C:\folder\file.txt", shell=True)

This works perfectly fine. The default editor is automatically opened and the file is loaded, however, I have read somewhere before that invoking calls through the shell (cmd.exe in Windows) is less safe. How can I do the same without it. Simply setting shell=False is giving me the error:

OSError: [WinError 193] %1 is not a valid Win32 application

Now, I can try this as a workaround:

subprocess.Popen("notepad C:\folder\file.txt")

but this only works if notepad is available, hence loses its generality.

Upvotes: 2

Views: 505

Answers (2)

Oliver Friedrich
Oliver Friedrich

Reputation: 9240

The feature you try to use is a builtin thing of the windows cmd.exe. Therefore you need to set the shell=True parameter. The cmd.exe knows what to do with the file you hand in.

If you use shell=False, you try to start the file like a programm and hence nothing happens, since .txt files have no exe-header.

Read more about it in the documentation.

Reading further, you can find why using the shell=True parameter can be a security flaw. If you consider to assemble the parameters by user inputs, you should not use this, otherwise nothing speaks against it.

Anyway, I recommend using your second example, because it is explicit. You decide what program to start.

subprocess.Popen("notepad C:\folder\file.txt")

Upvotes: 1

Simon Gibbons
Simon Gibbons

Reputation: 7184

If you are using windows then there is the (non portable) os.startfile command which will take a file path and open it in the default application for that filetype.

In your case you would do:

import os
os.startfile("C:\folder\file.txt")

Note that this method won't work on Linux and Mac OSX, you'll have to have to use their own utilities for this. (open for OSX and xdg-open on Linux) and start them with subprocess.

Upvotes: 2

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