How Can I Launch An Instance Of An Application Using Python?

January 18, 2024 Post a Comment

I am creating a Python script where it does a bunch of tasks and one of those tasks is to launch and open an instance of Excel. What is the ideal way of accomplishing that in my sc

Solution 1:

While the Popen answers are reasonable for the general case, I would recommend win32api for this specific case, if you want to do something useful with it:

It goes something like this:

from win32com.client import Dispatch
xl = Dispatch('Excel.Application')
wb = xl.Workbooks.Open('C:\\Documents and Settings\\GradeBook.xls')
xl.Visible = True# optional: if you want to see the spreadsheet

Taken from a mailing list post but there are plenty of examples around.

Solution 2:

os.system("start excel.exe <path/to/file>")

(presuming it's in the path, and you're on windows)

and also on Windows, just start <filename> works, too - if it's an associated extension already (as xls would be)

Solution 3:

The subprocess module intends to replace several other, older modules and functions, such as:

Baca Juga

os.system
os.spawn*
os.popen*
popen2.*
commands.*

import subprocess

process_one = subprocess.Popen(['gqview', '/home/toto/my_images'])

print process_one.pid

Solution 4:

I like popen2 for the ability to monitor the process.

excelProcess = popen2.Popen4("start excel %s" % (excelFile))
status = excelProcess.wait()

https://docs.python.org/2/library/popen2.html

EDIT: be aware that calling wait() will block until the process returns. Depending on your script, this may not be your desired behavior.

Solution 5:

As others have stated, I would suggest os.system. In case anyone is looking for a Mac-compatible solution, here is an example:

import osos.system("open /Applications/Safari.app")

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