when i try
os.system("open " + 'myfile.xlsx')
i get the output '0'
similarly, trying
os.system("start excel.exe myfilepath")
gives the result 32512
I have imported os and system, and I'm on mac. How can I change this so it does actually launch that excel file? And out of curiosity, what do the numbers it prints out mean?
Thanks!
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
Just these two lines
import os
os.system("start EXCEL.EXE file.xlsx")
Provided that file.xlsx is in the current directory.
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().
