Bash Script for deleting old backup files - bash

Question

every night my server creates backups for every mysql database. All of these files are saved in a folder like /backup/mysql/2014-11-28. Over the last few months a lot of folders and files have been stored in that directory and I would like to reduce this.
Therefore I would need a bash script which deletes every folder in the given directory, except every folder created this month (not in the last 30 days, but the actual month) and except one backup from every week (for example the backup from sunday). Since I have no clue how to do the sunday party I decided it would be simpler to keep the backups from the 07th, the 14th, the 21st and the the 28th.
#!/bin/bash
in_array() {
local haystack=${1}[#]
local needle=${2}
for i in ${!haystack}; do
if [[ ${i} == ${needle} ]]; then
return 0
fi
done
return 1
}
YEAR=`date +%Y`
LASTYEAR=`date +%Y -d "1 year ago"`
MONTH=`date +%m`
DAYS="07 14 21 28"
for i in $( ls ); do
BACKUP_YEAR=$(echo "${i}" | cut -d'-' -f1)
BACKUP_MONTH=$(echo "${i}" | cut -d'-' -f2)
BACKUP_DAY=$(echo "${i}" | cut -d'-' -f3)
DELETE=false
if [[ "$BACKUP_YEAR" == "$YEAR" ]]; then
if [[ "$BACKUP_MONTH" != "$MONTH" ]]; then
if [ $(in_array $DAYS "$BACKUP_DAY") ]; then
DELETE=true
fi
fi
fi
if [[ "$BACKUP_YEAR" == "$LASTYEAR" ]]; then
if [[ "$BACKUP_DAY" != ${DAYS[0]} ]]; then
DELETE=true
fi
fi
if [ "$DELETE" = true ] ; then
#OUTPUT=`rm -v -R $i`
echo $i
fi
done
The second part (backups from the year before) works perfectly fine, but the first part (for backups from this year) doesn't work and I don't know why. I found the function on this site, but I guess somehow my call is wrong.
EDIT: The code I am now using:
#!/bin/bash
read YEAR MONTH <<<$(date "+%Y %m")
LASTYEAR=$(( YEAR-1 ))
DAYS=" 07 14 21 28 "
for fn in $( ls )
do
if ([ "${fn:0:4}" = "$YEAR" ] &&
[ "${fn:5:2}" != "$MONTH" ] &&
[ "${DAYS/ ${fn:8:2} /}" = "$DAYS" ]) || ([ "${fn:0:4}" = "$LASTYEAR" ] &&
[ "${fn:8:2}" != ${DAYS:1:2} ])
then
#OUTPUT=`rm -v -R $fn`
echo "$fn"
fi
done

Answer 1

#!/bin/bash
read YEAR MONTH <<<$(date "+%Y %m")
LASTYEAR=$(( YEAR-1 ))
DAYS=" 07 14 21 28 "
for fn in $( ls )
do
DELETE=false
if [ "${fn:0:7}" = "$YEAR-$MONTH" ] &&
[ "${DAYS/ ${fn:8:2} /}" != "$DAYS" ]
then
DELETE=true
elif [ "${fn:0:4}" = "$LASTYEAR" ] &&
[ "${fn:8:2}" != ${DAYS:1:2} ]
then
DELETE=true
fi
if [ "$DELETE" = true ]
then
#OUTPUT=`rm -v -R $fn`
echo "$fn"
fi
done
Assuming that a file name matches the following date format YYYY-MM-DD, for instance, 2014-03-27, the expression ${fn:0:7} would be 2014-03; ${fn:8:2} would be 27; ${fn:0:4} would be 2014.
Read the bash manual pages about the parameter expansion.
The expression ${DAYS/ ${fn:8:2} /} checks the day from the filename (wrapped by white spaces) is found in the DAYS list. It would be equivalent to echo "${DAYS}" | grep -q "${BACKUP_DAY}".
note: it is possible to reduce the commands inside the for loop (join if and elif by means of a || and remove the DELETE variable checking and the DELETE variable itself) but I decided to keep the look of the original script.