I'm trying to open a excel so it is displayed visibly.
I am using:
import subprocess
subprocess.check_call(['open', '-a', 'C:\\1\\2\\22.xlsx'])
Sorry if this is obvious but I keep getting this error message. AS far as I can tell the file is located there.
FileNotFoundError: [WinError 2] The system cannot find the file specified
Not sure why
import os
os.path.join('C:', '1', '2', '22.xlsx')
Did not work. I think it worked but there was no visible excel.
Anyway, this works
xl=win32com.client.Dispatch("Excel.Application")
xl.Visible = True
xl.Workbooks.Open("C:\\1\\2\\22.xlsx")
it didn't work just because open isn't a known command name (this is an action in the windows registry, nothing more). The subprocess error message:
FileNotFoundError: [WinError 2] The system cannot find the file specified
has only one meaning: the executable could not be found. It doesn't (and cannot) check the parameters to see if they're valid filepaths so don't look in that direction (and yes, it's a pity that the message is generic and doesn't say exactly which file could not be found, that a known windows issue)
That would work:
import os
os.startfile(r'C:\1\2\22.xlsx')
os.startfile tries to perform the same thing as when you're clicking on the file in windows explorer.
This solution has the advantage not to depend on win32com.
Aside/discussed in comments: os.path.join('C:', '1', '2', '22.xlsx') doesn't work because it doesn't add the backslash after the drive colon (and for a "good" reason (well a good reason windows-wise): Python os.path.join on Windows).
>>> os.path.join('C:', '1', '2', '22.xlsx')
'C:1\\2\\22.xlsx' # wrong
a clean way to do it:
>>> os.path.join('C:', os.sep,'1' , '2', '22.xlsx')
'C:\\1\\2\\22.xlsx'
