Question

Using doctl in bash script

I have this script so far:

#!/bin/bash
# script to create digitalocean droplet snapshot using their CLI, doctl.

# Function to set variable for snapshot name as date/time of creation.
timestamp() {
NAME=$(date)
}

timestamp

doctl compute droplet-action snapshot --snapshot-name "$NAME" 80750079

The script works with the result being a droplet snapshot with the date/time of creation as the name. However, I need to keep only 12 snapshots.

Somehow I need to count the number of existing snapshots and when I have 12, delete the oldest and add a new one.

How can I achieve that?

I run the script every two hours.

Any help would be greatly appreciated!

Thanks!


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So this is the bashscript. Uses doctl to check how many snapshots exist, and when 24 exist, deletes the oldest one and creates the new one. Sends an log email to me with each new snapshot so I know it’s working. Repeats every hour. Maybe it’ll help someone. :)

#!/bin/bash
exec &> do_snapshot.log

timestamp() {
 NAME=$(date)
}

timestamp

SNAPSHOTS=$(/snap/bin/doctl compute image list-user --format "ID" --no-header | wc -l)

if [ "$SNAPSHOTS" -gt 23 ]; then
mapfile -t IDLIST < <(/snap/bin/doctl compute image list-user --format "ID" --no-header)

OLDEST=${IDLIST[0]}

/snap/bin/doctl compute image delete "$OLDEST"
fi

/snap/bin/doctl compute droplet-action snapshot --snapshot-name "$NAME" 8xxxxxxxxx9

mail -s "Digital Ocean Hourly Snapshot" my email@gmail.com < do_snapshot.log

rm do_snapshot.log

exit 0

BTW, I’ll post the script when completed.

Ryan Quinn
DigitalOcean Employee
DigitalOcean Employee badge
April 23, 2018

I think that if you list your snapshots and split that result so that each result is a different item in an array you could do a for loop to count your results and if that number is over 12, delete the oldest.

Personally I would choose a more robust option than bash+doctl for this since it would be easier to deal with the snapshot results if you were using a library/class rather than a command line utility.

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