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?
7 Answers
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.
1 Comment
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)
1 Comment
os.system(). IMHO it's better to use os.startfile().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.
Comments
os.system("open file.xls")