How can I launch an instance of an application using Python?
Question
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 script?
Solution
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.
OTHER TIPS
or
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)
I like popen2 for the ability to monitor the process.
excelProcess = popen2.Popen4("start excel %s" % (excelFile))
status = excelProcess.wait()
http://www.python.org/doc/2.5.2/lib/module-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.
The subprocess module intends to replace several other, older modules and functions, such as:
- os.system
- os.spawn*
- os.popen*
- popen2.*
- commands.*
.
import subprocess
process_one = subprocess.Popen(['gqview', '/home/toto/my_images'])
print process_one.pid
As others have stated, I would suggest os.system. In case anyone is looking for a Mac-compatible solution, here is an example:
import os
os.system("open /Applications/Safari.app")
os.system("open file.xls")
I like os.startfile("path to file")
as it opens the file as if you've double clicked to open.
I found that with os.system("start excel filename")
it opened it like a file opened from the web and you had to enable editing.