Unix - how to fetch files based on the part of a filename that is date

Assuming you have GNU date, this function will get you all the dates for the previous 2 months:

dates_in_previous_two_months () { 
    for month_delta in 0 1; do
        # get the last day of a previous month
        day=$(date -d "$(date +%Y-%m-01) -$month_delta month -1 day" +%Y%m%d)
        until [[ $day == *00 ]]; do
            echo $day
            ((day--))
        done
    done
}

For any day in the month of February 2013, this returns:

20130131
20130130
20130129
[... and so on until ...]
20121203
20121202
20121201

In March it will return dates from 20130228 to 20130101

You could use it like:

for date in $(dates_in_previous_two_months); do
    "fetch" AA_XX_$date
done

or

dates_in_previous_two_months | xargs -I{} "fetch" AA_XX_{}

You can use find command. The following command will list down all files modified within last 60 days

find . -name 'AA_XX_*' -mtime -60

Here is shell script for achieving what you want, but I would agree that it's not a clean solution. I, myself, don't like this clumsy code, but then it will work for you, for time being

#!/usr/local/bin/bash

YEAR=`date +%Y`
MONTH=`date +%m`
DATE=`date +%d`

if [ $MONTH = 01 ]; then
    OLDEST=`expr $YEAR - 1`11010000
    NEWEST=`expr $YEAR - 1`12312359
elif [ $MONTH = 02 ]; then
    OLDEST=`expr $YEAR - 1`12010000
    echo $YEAR
    NEWEST="$YEAR"01312359
else
    case $MONTH in
            03)
            OLDEST=$YEAR0`expr $MONTH - 2`010000
            if [ $YEAR%4 = 0 ]; then
                    DATE=29
            else
                    DATE=28
            fi
            NEWEST=$YEAR0`expr $MONTH - 1`"$DATE"2359;;

            04|06|09|11)
            OLDEST=$YEAR0`expr $MONTH - 2`010000
            NEWEST=$YEAR0`expr $MONTH - 1`312359;;

            07|08|10|12)
            OLDEST=$YEAR0`expr $MONTH - 2`010000
            NEWEST=$YEAR0`expr $MONTH - 1`302359;;
        esac
fi

echo $OLDEST
echo $NEWEST

touch -t $OLDEST oldest
touch -t $NEWEST newest

find . -name 'AA_XX_*' -newer oldest ! -newer newest > temp.txt

Like this?

cp source/AA_XX_$(date -d "2 months ago" +%04Y%02m)* source/AA_XX_$(date -d "1 month ago" +%04Y%02m)* destination/

This month with the year prepended is $(date +%Y%m). Last month is previous_month_1=$(date -d '1 month ago' +%Y%m) with GNU date (i.e. on non-embedded Linux or Cygwin), and the month before is previous_month_2=$(date -d '2 months ago' +%Y%m). on other implementations, they're given by

this_month=$(date +%Y%m)
case $this_month in
  *01) previous_month_1=$((this_month-89)); previous_month_2=$((previous_month-1));;
  *02) previous_month_1=$((this_month-1)); previous_month_2=$((previous_month-89));;
  *) previous_month_1=$((this_month-1)); previous_month_2=$((previous_month-1));;
esac

Now, to match all existing files dated from the previous two months:

fetch "AA_XX_$previous_month_1"?? "AA_XX_$previous_month_2"??

(Replace fetch by the command you want to run on those files.) If a month may have no matching file, the pattern will be left unchanged:

filter_month () {
  case $1 in *'??') return;; esac
  fetch "$@"
}
filter_month "$previous_month_1"??
filter_month "$previous_month_2"??

If you want to fetch all files corresponding to the days of that month, you'll either need some GNU date magic (not described here) or some leap year logic.

filter_month () {
  case $1 in
    *0[13578]|*1[02]) n=31;;
    *0[469]|*11) n=30;;
    *0002) if [ $((${1%????} %4)) -eq 0 ]; then n=29; else n=28; fi;;
    *02) if [ $((${1%??} %4)) -eq 0 ]; then n=29; else n=28; fi;;
  esac
  while [ $n -gt 0 ]; do
    d=$n
    if [ $d -lt 10 ]; then d=0$d; fi
    fetch "AA_XX_$1$d"
  fi
}
filter_month "$previous_month_1"
filter_month "$previous_month_2"