statvfs

[Marc 'BlackJack' Rintsch]

On Mon, 29 Oct 2007 16:52:12 -0500, Korthrun wrote:

> I'm writing some scripts to populate RRD's, mainly for practicing python.
> 
> As such I've decided to play with statvfs in order to build disk
> graphs. Here is what I have and what I get. What I'm curious about
> here is the meaning of the "L" charcter, as that fubars math.

The L means its a `long` literal, i.e. an integer value that can be
arbitrary big.  Well its limited by memory of course.

And it doesn't "fubar" math, the L just shows up in the string
representation but you don't calculate with strings but numbers.

> ###start code###
> 
> from os import statvfs, path
> from statvfs import *
> 
> mps = [ '/', '/home', '/www', '/var', '/storage/backups',
> '/storage/snd' ]
> 
> for fs in mps:
> 	data = statvfs(fs)
> 	print data
> 
> ###end code, start output###
> (4096, 4096, 4883593L, 4045793L, 4045793L, 0L, 0L, 0L, 1024, 255)
> (4096, 4096, 1220889L, 1114718L, 1114718L, 0L, 0L, 0L, 0, 255)
> (4096, 4096, 19267346L, 18273138L, 18273138L, 0L, 0L, 0L, 0, 255)
> (4096, 4096, 3662687L, 3492397L, 3492397L, 0L, 0L, 0L, 0, 255)
> (4096, 4096, 3417702L, 2116063L, 2116063L, 0L, 0L, 0L, 0, 255)
> (4096, 4096, 25885944L, 21799115L, 21799115L, 0L, 0L, 0L, 0, 255)
> ###end output###
> 
> Ideally I'll just do some blocksize * number of blocks math to get
> the % of used space.

Just go ahead and do it:

In [185]: stat = os.statvfs('/')

In [186]: stat.f_bsize
Out[186]: 4096

In [187]: stat.f_blocks
Out[187]: 2622526L

In [188]: stat.f_bsize * stat.f_blocks
Out[188]: 10741866496L

Ciao,
	Marc 'BlackJack' Rintsch