How to do date calculations in Shell Scripting?

Question

I have a shell script that runs every night to backup my EC2 sites database and html to S3, and when it backs the folders up, it appends the date to it for easier viewing. But I want it to also be able to delete the relevant backup folders from 3 days before. How can I do the calculations to get the date 3 days ago?

#!/bin/bash
DATE=`date +%m%d%Y`

s3cmd put -r /var/lib/mysql/mydb/ s3://mybucket/mydb-$DATE/
s3cmd put -r /home/ec2-user/public_html/ s3://mybucket/public_html-$DATE/
s3cmd del -r s3://mybucket/mydb-(date 3 days ago)

Karonkaross

Answer 1

You can use the -d flag for the date command:

-d, --date=STRING
     display time described by STRING, not 'now'

So, just change your date variable to:

DATE=`date +%m%d%Y -d "3 days ago"`

Gemma

Thanks! I thought I need to use awk or something. – Caravaggio

And here is the list of recognized formats: gnu.org/software/coreutils/manual/html_node/… – Prager

Answer 2

Why don't you use the modification time of the directories? Then you can just search for them with find. For exmaple:

find backups -maxdepth 1 -mtime 3

Gebhardt

Very helpful!, use -mmin -60 to look for files modified within the last 60 minutes. – Halter

Answer 3

It's different on BSD / Mac. You need to use the -v option:

date -v -3d +%m%d%Y

Belovo