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Top Forums UNIX for Beginners Questions & Answers [Tip] Housekeeping Tasks Made Easy - User Home directories and Leftover Files Post 303037287 by bakunin on Monday 29th of July 2019 08:45:16 AM
Old 07-29-2019
Quote:
Originally Posted by MadeInGermany
1. scanning "unowned" homedirs for recently accessed files. If nothing found, delete. If something found, display it and stop the search - and do not delete.
Yes, that is another possible solution. A problem could be that users put things in their homedir crontab and so some files get regularly accessed even if the accounts are deleted. If this or my solution is better is perhaps depending on the environment you work in, policies in place and - last but not least - personal taste. The real point, though, is to take care of (removed users) data in some way in specific and to not let accumulate data waste on the system in general.

Quote:
Originally Posted by MadeInGermany
2. scanning shared project directories in "deepest first fashion" (find -depth), and assign each "unowned" directory to the owner of its parent directory.
This is a very good idea! I will update the above script eventually when i find time.

I hope this helps.

bakunin
 

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SHELL-QUOTE(1)						User Contributed Perl Documentation					    SHELL-QUOTE(1)

NAME
shell-quote - quote arguments for safe use, unmodified in a shell command SYNOPSIS
shell-quote [switch]... arg... DESCRIPTION
shell-quote lets you pass arbitrary strings through the shell so that they won't be changed by the shell. This lets you process commands or files with embedded white space or shell globbing characters safely. Here are a few examples. EXAMPLES
ssh preserving args When running a remote command with ssh, ssh doesn't preserve the separate arguments it receives. It just joins them with spaces and passes them to "$SHELL -c". This doesn't work as intended: ssh host touch 'hi there' # fails It creates 2 files, hi and there. Instead, do this: cmd=`shell-quote touch 'hi there'` ssh host "$cmd" This gives you just 1 file, hi there. process find output It's not ordinarily possible to process an arbitrary list of files output by find with a shell script. Anything you put in $IFS to split up the output could legitimately be in a file's name. Here's how you can do it using shell-quote: eval set -- `find -type f -print0 | xargs -0 shell-quote --` debug shell scripts shell-quote is better than echo for debugging shell scripts. debug() { [ -z "$debug" ] || shell-quote "debug:" "$@" } With echo you can't tell the difference between "debug 'foo bar'" and "debug foo bar", but with shell-quote you can. save a command for later shell-quote can be used to build up a shell command to run later. Say you want the user to be able to give you switches for a command you're going to run. If you don't want the switches to be re-evaluated by the shell (which is usually a good idea, else there are things the user can't pass through), you can do something like this: user_switches= while [ $# != 0 ] do case x$1 in x--pass-through) [ $# -gt 1 ] || die "need an argument for $1" user_switches="$user_switches "`shell-quote -- "$2"` shift;; # process other switches esac shift done # later eval "shell-quote some-command $user_switches my args" OPTIONS
--debug Turn debugging on. --help Show the usage message and die. --version Show the version number and exit. AVAILABILITY
The code is licensed under the GNU GPL. Check http://www.argon.org/~roderick/ or CPAN for updated versions. AUTHOR
Roderick Schertler <roderick@argon.org> perl v5.16.3 2010-06-11 SHELL-QUOTE(1)
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