CODEWITHSUNDEEP
arraysbashutc
user2642953
user2642953

Reputation: 55

Comparing UTC date/time strings in Bash

I have an array that contains the UTC creation times for a few amazon ec2 volume snapshots that are in this format: 2013-08-09T14:20:47.000Z

I'm trying to figure out a way to compare the elements of the array to each other to determine which one is the oldest snapshot and delete it in Bash 4

The current code that I have right now:

#creates a count of all the snapshot volume-id's and deletes oldest snapshot if 
#there are more than 5 snapshots of that volume

declare -A vol_id_count=( ["id_count"]="${snapshot_vols[@]}" )
check_num=5

for e in ${vol_id_count[@]}
do
  if (( ++vol_id_count[$e] > $check_num ))
  then
    echo "first nested if works"
    #compare UTC times to find oldest snapshot
    #snapshot_time=${snapshot_times[0]}

    #for f in ${snapshot_times[@]}
    #do
    #  time= date --date="$snapshot_time" +%s
    #  snapshot_check=${snapshot_times[$f]}
    #  echo "check: "$snapshot_check
    #  check= date --date="$snapshot_check" +%s

    #  if [[ "$snapshot_time" -gt "$snapshot_check" ]]
    #  then
    #    snapshot_time=$snapsnapshot_check
    #    echo "time: "$snapshot_time
    #  fi
    #done

    #snapshot_index=${snapshot_times[$snapshot_time]}
    #aws ec2 delete-snapshot --snapshot-id "${snapshot_ids[$snapshot_index]}"
  fi
done

I have the first for loop and if statement working to check if there are more than 5 snapshots of a certain volume but I'm scratching my head trying to figure out how to compare the UTC strings. Would a second associative array do the trick I wonder?

Upvotes: 1

Views: 979

Answers (2)

aks
aks

Reputation: 2346

If you already have an array, no need to use a loop to print the items. A single echo will work (but only if there are no spaces in the filenames):

oldest=`echo "${snapshot_times[@]}" | tr ' ' '\12' | sort -r | tail -1)`

If those "timestamps" are coming from the filenames themselves, just use ls in a pipe:

snapshot_dir=/some/path/to/snapshots
count=`ls -1 $snapshot_dir/ | wc -l`  # get count of snapshots

# expunge the oldest snapshot if there are more than 5
if (( count > 5 )); then
  oldest_name=`(cd $snapshot_dir ; ls -1 ) | sort | head -1`
  oldest_file="$snapshot_dir/$oldest_name"
  [[ -f "$oldest_file" ]] && {
    echo "Purging oldest snapshot: $oldest_file"
    rm -f $oldest_file
  }
fi

Upvotes: 0

pgl
pgl

Reputation: 8021

Here's how I would do it: with sort.

dates=(
    2013-08-09T14:20:47.000Z
    2013-08-09T14:31:47.000Z
    )

latest=$(for date in ${dates[@]}
do
    echo $date
done | sort | tail -n 1)

echo $latest # outputs "2013-08-09T14:31:47.000Z"

This is obviously just a sample script, modify as needed - it could easily be turned into a function like:

function latest() {
    local dates="$@"
    for date in ${dates[@]}
    do
        echo $date
    done | sort | tail -n 1
}

latest ${dates[@]} # outputs "2013-08-09T14:31:47.000Z"

Upvotes: 2

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