Get the date before the provided date in Bash

Question

I have a bash script that receives a date as a parameter ($1) in the format YYYYMMDD

What I need to do is subtract one day from the provided date. Lets take 20130701 as an example, after processing I expect to get 20130630.

How can I do this in bash? (One line solution gets bonus points!)

glenn jackman · Accepted Answer · 2013-07-03 19:02:36Z

6

Assuming you have GNU date:

$ set -- 20130701 $ yesterday=$(date -d "$1 - 1 day" "+%Y%m%d") $ echo "$yesterday" 20130630

answered Jul 3, 2013 at 19:02

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