Date: Sat, 2 Apr 2011 09:20:27 -0500 From: Ryan Coleman <editor@d3photography.com> To: FreeBSD Mailing List <freebsd-questions@freebsd.org> Subject: graphical representation of `du` Message-ID: <0DD2BF5C-7387-4AFA-BF43-B1683F3773E8@d3photography.com>
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I found this command: ls -R | grep ":$" | sed -e 's/:$//' -e 's/[^-][^\/]*\//--/g' -e 's/^/ /' -e 's/-/|/' Which makes this: |-Mar17 |---1300074369-chow |-----download |-------small |---1300421616-Cunningham |-----download |-------small But I want to use `du` instead to convert this 2.0M ./Mar17/1300074369-chow/download/small 2.0M ./Mar17/1300074369-chow/download 2.0M ./Mar17/1300074369-chow 2.1M ./Mar17/1300421616-Cunningham/download/small 2.1M ./Mar17/1300421616-Cunningham/download 2.1M ./Mar17/1300421616-Cunningham 4.1M ./Mar17 into this: |-Mar17 [4.3M] |---1300074369-chow [2.0M] |-----download [2.0M] |-------small [2.0M] |---1300421616-Cunningham [2.1M] |-----download [2.1M] |-------small [2.1M] I realize it does it backwards and I can live with that... OR mix the two to run the first command and run another command to get the folders total size or something... you know? Thanks for the help, Ryanhome | help
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