[Python-Dev] Function in os module for available disk space, why not?

[M.-A. Lemburg]

Fredrik Lundh wrote:
> 
> dinu wrote:
> > Well, this is the usual "If you need it, do it yourself!"
> > answer, that bites the one who dares to speak up for all
> > those hundreds who don't... isn't it?
> 
> fwiw, Python already supports this for real Unix platforms:
> 
> >>> os.statvfs("/")
> (8192, 512, 524288, 365788, 348336, 600556, 598516, 598516, 0, 255)
> 
> here, the root disk holds 524288x512 bytes, with 348336x512
> bytes free for the current user, and 365788x512 bytes available
> for root.
> 
> (the statvfs module contains indices for accessing this "struct")
> 
> Implementing a small subset of statvfs for Windows wouldn't
> be that hard (possibly returning None for fields that don't make
> sense, or are too hard to figure out).
> 
> (and with win32all, I'm sure it can be done without any C code).

It seems that all we need is Jack to port this to the Mac
and we have a working API here :-)

Let's do it...

import sys,os

try:
    os.statvfs

except KeyError:
    # Win32 implementation...
    # Mac implementation...
    pass

else:
    import statvfs
    
    def freespace(path):
        """ freespace(path) -> integer
        Return the number of bytes available to the user on the file system
        pointed to by path."""
        s = os.statvfs(path)
        return s[statvfs.F_BAVAIL] * long(s[statvfs.F_BSIZE])

if __name__=='__main__':
    path = sys.argv[1]
    print 'Free space on %s: %i kB (%i bytes)' % (path,
                                                  freespace(path) / 1024,
                                                  freespace(path))

-- 
Marc-Andre Lemburg
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