CODEWITHSUNDEEP
pythonsubprocess
Evan Gertis
Evan Gertis

Reputation: 2062

How to use Popen to open a program in Windows

I am working on a Python project to open an application in Windows. The following program is written like so

import subprocess
subprocess.Popen(['open','C:\\Windows\\System32\\calc.exe'])

Expected: The calculator app should open.

Actual:

  File "C:\Program Files\WindowsApps\PythonSoftwareFoundation.Python.3.10_3.10.1520.0_x64__qbz5n2kfra8p0\lib\subprocess.py", line 969, in __init__
    self._execute_child(args, executable, preexec_fn, close_fds,
  File "C:\Program Files\WindowsApps\PythonSoftwareFoundation.Python.3.10_3.10.1520.0_x64__qbz5n2kfra8p0\lib\subprocess.py", line 1438, in _execute_child
    hp, ht, pid, tid = _winapi.CreateProcess(executable, args,
FileNotFoundError: [WinError 2] The system cannot find the file specified
PS C:\Users\EvanGertis\development\PythonAutomation\Module9>

Upvotes: 0

Views: 1453

Answers (1)

Cardstdani
Cardstdani

Reputation: 5223

Removing open parameter argument from the function Popen works well in Python 3:

import subprocess
subprocess.Popen(['C:\\Windows\\System32\\calc.exe'])

Also, you can use call() instead of Popen(), with the same result:

subprocess.call(['C:\\Windows\\System32\\calc.exe'])

Upvotes: 2

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