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Home/ Questions/Q 8861651
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Editorial Team
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Editorial Team
Asked: June 14, 20262026-06-14T15:32:19+00:00 2026-06-14T15:32:19+00:00

I am trying to create a script which grabs the most recently dated file

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I am trying to create a script which grabs the most recently dated file from an S3 amazon bucket. This is the code I am working with at the moment:

#!/usr/bin/env python

import boto
import boto.s3
from boto.s3.key import Key

# Define variables
AWS_ACCESS_KEY_ID = ''
AWS_SECRET_ACCESS_KEY = ''
BUCKET_NAME = 'My Bucket'
BUCKET = conn.get_bucket(bucket_name)
CONN = boto.connect_s3(AWS_ACCESS_KEY_ID, AWS_SECRET_ACCESS_KEY)

for key in bucket.list():
    print key.name

I havent had chance to sign up to amazon to test this yet, but I found this code on the internet and it should list all the files in the bucket.

Each of my files are dated with the following name format:

NOW = datetime.datetime.now().strftime("%Y-%m-%d__%H-%M")
Example: 1999-01-30__10-30 (10:30 on the 30th Jan 1999)

My question is, if I wanted to work out which is the most recent file based on its dated filename, what is a good way of going about this?

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  1. Editorial Team
    Editorial Team
    2026-06-14T15:32:20+00:00Added an answer on June 14, 2026 at 3:32 pm

    Just sort the filenames; the most recent file will be the last one:

    >>> sorted(['1999-01-30__10-30', '2012-10-10__11-22', '2011-01-30__22-11'])
    ['1999-01-30__10-30', '2011-01-30__22-11', '2012-10-10__11-22']
    
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