Date: Fri, 18 Jan 2013 18:51:11 -0500 (EST) From: Chris Hill <chris@monochrome.org> To: Fbsd8 <fbsd8@a1poweruser.com> Cc: FreeBSD questions <questions@freebsd.org> Subject: Re: sh script code to get file size. Message-ID: <alpine.BSF.2.00.1301181846160.42367@tripel.monochrome.org> In-Reply-To: <50F9DA3E.5050607@a1poweruser.com> References: <50F9DA3E.5050607@a1poweruser.com>
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On Fri, 18 Jan 2013, Fbsd8 wrote: > In a script in am working on I need to find out the allocated > size of a sparse file. > The only command that comes to mind is "ls -lh" > The "du -h" command is not appropriate because it will show > the occupied size and not the allocated size. > > I don't know how to parse out to the position in the output of that > "ls -lh" command to pickup the file size value. > > Is there some other way to do this? To parse it out, I've used something like: $ ls -lh npviewer.bin.core | cut -d \ -f 9 186M After the backslash are two spaces: one being the space that's being escaped to make it the delimiter, the other to separate the options. The number after the '-f' determines which "field" of the output is displayed, which may vary. HTH. -- Chris Hill chris@monochrome.org ** [ Busy Expunging </> ]
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