Opening/running Excel file from python
Question: I need to start excel and open a file directly from python. Currently I am
using:

    
    
    import os
    os.system('start excel.exe file.xls')
    

However I do not get desired result. I want to open a file from local
destination (file is in the same folder with program), but this code open's
the file with same name in my home (user) directory and not from my program
directory.


Answer: The problem is that the directory where the program is located is not used.
The current working directory is. So you have to find out which directory your
program is located in, which python conveniently prepared for you in:

    
    
    sys.path[0]
    

and either change directory to it:

    
    
    os.chdir(sys.path[0])
    

or give full path for the file you want to open

    
    
    os.system('start excel.exe "%s\\file.xls"' % (sys.path[0], ))
    

Note, that while Windows generally accept forward slash as directory
separator, the command shell (`cmd.exe`) does not, so backslash has to be used
here. `start` is Windows-specific, so it's not big problem to hardcode it
here. More importantly note that Windows don't allow `"` in file-names, so the
quoting here is actually going to work (quoting is needed, because the path is
quite likely to contain space on Windows), **but it's bad idea to quote like
this in general!**

