Use Python to launch Excel file

Question

when i try os.system(\"open \" + \'myfile.xlsx\')
i get the output \'0\'

similarly, trying
os.system(\"start excel.exe myfil

Answer 1

I don't know about Mac OS, but if Windows Operating System is the case and provided the Microsoft Windows is properly installed, then consider using :

import os
os.system('start "excel" "C:\\path\\to\\myfile.xlsx"')

double-quotation is important for excel and C:\\path\\to\\myfile.xlsx ( where C just denotes the letter for the partition within the file system, might be replaced by D,E ..etc. ), and single-quotation is needed for the whole string within the os.system().

Answer 2

On Windows 10, this works for me:

import os
full_path_to_file = "C:\blah\blah\filename.xlsx"
os.system(full_path_to_file)

Copy-pasting any full path to a file in the command prompt (or passing it to os.system()) seems to work as long as you have permission to open the file. I suppose this only works when Excel is selected as default application for the .xlsx extention.

Answer 3

If you only want to open the excel application you could use subprocess:

import subprocess
subprocess.check_call(['open', '-a', 'Microsoft Excel'])

You can also use os and open a specific file:

import os
os.system("open -a 'path/Microsoft Excel.app' 'path/file.xlsx'")

If you on other hand want to open an excel file within python and modify it there's a number of packages to use as xlsxwriter, xlutils and openpyxl where the latter is prefered by me.

Another note, if you're on mac the excel application isn't .exe

Answer 4

import os

os.startfile(dirpath + '\\' + 'filename.xlsx')

Answer 5

Just these two lines

import os
os.system("start EXCEL.EXE file.xlsx")

Provided that file.xlsx is in the current directory.