问题
Amazon "s3cmd ls" takes like this output:
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 files of archive, ending at 00 ... XX, with the latest date of fileset ?
Date and time is not sorted.
Bash, regexp ?
Thanx!
回答1:
DATE=$(s3cmd ls | sort -n | tail -n 1 | awk '{print $1}')
s3cmd ls | grep $DATE
sorting as a number schould put youngest dates last. Tail -n1 takes the last line, awk cuts the first word which is the date. Use that to get all entries of that date.
But maybe I didn't understand the question - so you have to rephrase it. You tell us 'date and time is not sorted' and provide an example where they are sorted - you ask for the latest date, but all entries have the same date.
回答2:
s3cmd ls s3://bucket/path/ | sort -k1,2
This will sort by date ascending.
回答3:
Try
This command syntax is s3 ls s3:// path of the bucket / files | sort
s3cmd ls s3://file1.tgz.00<br> | sort
It will sort in the descending date at last
来源：https://stackoverflow.com/questions/2674117/how-to-sort-output-of-s3cmd-ls