How to retrieve the most recent file in cloud storage bucket? - google-cloud-platform

Question

Is this something that can be done with gsutil?
https://cloud.google.com/storage/docs/gsutil/commands/ls does not seem to mention any sorting functionality - only filtering by a date - which wouldn't work for my use case.

Answer 1

Hello this still doesn't seems to exists, but there is a solution in this post: enter link description here
The command used is this one:
gsutil ls -l gs://[bucket-name]/ | sort -k 2
As it allow you to filter by date you can get the most recent result in the bucket and recuperating the last line using another pipe if you need.

Answer 2

gsutil ls -l gs://<bucket-name> | sort -k 2 | tail -n 2 | head -1 | cut -d ' ' -f 7
It will not work well if there is less then two objects in the bucket though

Answer 3

By using gsutil from a host machine this will populate the response array:
response=(`gsutil ls -l gs://some-bucket-name|sort -k 2|tail -2|head -1`)
Or by gsutil from docker container:
response=(`docker run --name some-container-name --rm --volumes-from gcloud-config -it google/cloud-sdk:latest gsutil ls -l gs://some-bucket-name|sort -k 2|tail -2|head -1`)
Afterwards, to get the whole response, run:
echo ${response[#]}
will print for example:
33 2021-08-11T09:24:55Z gs://some-bucket-name/filename-37.txt
Or to get separate info from the response, (e.g. filename)
echo ${response[2]}
will print the filename only
gs://some-bucket-name/filename-37.txt

Answer 4

For my use case, I wanted to find the most recent directory in my bucket. I number them in ascending order (with leading zeros), so all I need to get the most recent one is this:
gsutil ls -l gs://[bucket-name] | sort | tail -n 1 | cut -d '/' -f 4
list the directory
sort alphabetically (probably unnecessary)
take the last line
tokenise it with "/" delimiter
get the 4th token, which is the directory name