How to sort the output of "s3cmd ls"

Question

Amazon "s3cmd ls" takes this result:

2010-02-20 21:01 1458414588 s3://file1.tgz.00<br> 2010-02-20 21:10 1458414527 s3://file1.tgz.01<br> 2010-02-20 22:01 1458414588 s3://file2.tgz.00<br> 2010-02-20 23:10 1458414527 s3://file2.tgz.01<br> 2010-02-20 23:20 1458414588 s3://file2.tgz.02<br> <br> 

How to select all archive files ending in 00 ... XX with the latest file set date?

Date and time are not sorted.

Bash, regexp?

Thanx!

Answer 1

 DATE=$(s3cmd ls | sort -n | tail -n 1 | awk '{print $1}') s3cmd ls | grep $DATE 

sorting as a number can lead to the earliest dates. Tail -n1 takes the last line, awk cuts the first word, which is the date. Use this to get all entries for this date.

But maybe I did not understand the question, so you need to rephrase it. You tell us that “date and time are not sorted” and give an example of their sorting - you request the latest date, but all records have the same date.

Answer 2

 s3cmd ls s3://bucket/path/ | sort -k1,2 

It will sort by date in ascending order.