92 lines
2.5 KiB
Text
92 lines
2.5 KiB
Text
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#!/bin/sh
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# mdate-sh - get modification time of a file and pretty-print it
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# Copyright (C) 1995 Software Foundation, Inc.
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# Written by Ulrich Drepper <drepper@gnu.ai.mit.edu>, June 1995
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#
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# This program is free software; you can redistribute it and/or modify
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# it under the terms of the GNU General Public License as published by
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# the Free Software Foundation; either version 2, or (at your option)
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# any later version.
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#
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# This program is distributed in the hope that it will be useful,
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# but WITHOUT ANY WARRANTY; without even the implied warranty of
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# MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
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# GNU General Public License for more details.
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#
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# You should have received a copy of the GNU General Public License
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# along with this program; if not, write to the Free Software
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# Foundation, Inc., 59 Temple Place - Suite 330, Boston, MA 02111-1307, USA.
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# Prevent date giving response in another language.
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LANG=C
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export LANG
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LC_ALL=C
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export LC_ALL
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LC_TIME=C
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export LC_TIME
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# Get the extended ls output of the file.
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if ls -L /dev/null 1>/dev/null 2>&1; then
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set - `ls -L -l $1`
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else
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set - `ls -l $1`
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fi
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# The month is at least the fourth argument.
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# (3 shifts here, the next inside the loop)
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shift
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shift
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shift
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# Find the month. Next argument is day, followed by the year or time.
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month=
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until test $month
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do
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shift
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case $1 in
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Jan) month=January; nummonth=1;;
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Feb) month=February; nummonth=2;;
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Mar) month=March; nummonth=3;;
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Apr) month=April; nummonth=4;;
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May) month=May; nummonth=5;;
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Jun) month=June; nummonth=6;;
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Jul) month=July; nummonth=7;;
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Aug) month=August; nummonth=8;;
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Sep) month=September; nummonth=9;;
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Oct) month=October; nummonth=10;;
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Nov) month=November; nummonth=11;;
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Dec) month=December; nummonth=12;;
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esac
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done
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day=$2
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# Here we have to deal with the problem that the ls output gives either
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# the time of day or the year.
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case $3 in
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*:*) set `date`; eval year=\$$#
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case $2 in
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Jan) nummonthtod=1;;
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Feb) nummonthtod=2;;
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Mar) nummonthtod=3;;
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Apr) nummonthtod=4;;
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May) nummonthtod=5;;
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Jun) nummonthtod=6;;
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Jul) nummonthtod=7;;
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Aug) nummonthtod=8;;
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Sep) nummonthtod=9;;
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Oct) nummonthtod=10;;
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Nov) nummonthtod=11;;
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Dec) nummonthtod=12;;
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esac
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# For the first six month of the year the time notation can also
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# be used for files modified in the last year.
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if (expr $nummonth \> $nummonthtod) > /dev/null;
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then
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year=`expr $year - 1`
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fi;;
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*) year=$3;;
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esac
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# The result.
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echo $day $month $year
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