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!
Just these two lines
import os
os.system("start EXCEL.EXE file.xlsx")
Provided that file.xlsx is in the current directory.
os.system('start excel.exe file.xlsx') will open file with same name from your documents folder and os.system('start "excel.exe" "file.xlsx"') (note the extra quotes) will open file from same folder as code is in. Testing this on my end I am not sure y this happens but probably an important distinction.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.
full_path_to_file could contain blank character(s) / space(s), better add double quotes before and after full_path_to_file like: os.system('"' + full_path_to_file + '"')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
a parameter do? Thank you.
os.system("open -a 'Microsoft Excel' 'path/file.xlsx'") and this works too: subprocess.check_call(['open', '-a', 'Microsoft Excel','path/file.xlsx'])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().
As in: How to open an Excel file with Python to display its content?
import os
file = "C:\\Documents\\file.xlsx"
os.startfile(file)
It opens the file with the default application.
A minor improvement that handles spaces and explicitly uses Excel:
import os
file = "C:/Documents/file name with spaces.xlsx"
file_with_quotes = '"' + file + '"'
os.system('start "EXCEL.exe" {}'.format(file_with_quotes))
If you want this to work with your most recent downloaded xlsx file, you can use the following:
import glob
import os
list_of_files = glob.glob('C:\\Downloads\\*.xlsx') *#note the .xlsx extension is needed to retrieve the last downloaded xlsx file.*
latest_file = max(list_of_files, key=os.path.getctime)
os.system(latest_file)
0means success, and any other number means failure. Often themanpage for a program will tell you a list of the failure codes and what they mean.