For example, I have date: 4 August 1993 and I want to add 348 days to it, how can I do it in bash?
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3 Answers
Just use the date command with -d option:
$ date -d "1983-08-04 348 days"
Tue Jul 17 00:00:00 BST 1984
You can change the output format if you want:
$ date -d "1983-08-04 2 days" +%Y-%m-%d
1983-08-06
answered Sep 23, 2009 at 13:54
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1You can use the OP's date format, too:
date -d "4 August 1993 348 days" +"%d %B %Y"Sep 23, 2009 at 16:54 -
2
In bash on Mac OS X, you can do this:
date -j -v +348d -f "%Y-%m-%d" "1993-08-04" +%Y-%m-%d
Output: 1994-07-18
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3Been looking for this for a while. I appreciate. I wanted to replace the number "348" with a variable from a bash script. I ended up with
NEXT_DATE=$(date -j -v +$(( incrementDays ))d -f "%Y-%m-%d" "1993-08-04" +%Y-%m-%d)for anyone else looking to do this.– Ian GJan 15, 2019 at 21:11 -
Nice! There's a whole slew of formatting options you can use with the Bash
datecommand at tutorialkart.com/bash-shell-scripting/…– AFKSep 18, 2021 at 3:56 -
You can't believe how many incorrect answers I've stumbled upon until this one. Thanks!– MatthewOct 6, 2022 at 18:03
Here is a little more complex usage of this:
for i in `seq 1 5`;
do;
date -d "2014-02-01 $i days" +%Y-%m-%d;
done;
or with pipes:
seq 1 5 | xargs -I {} date -d "2014-02-01 {} days" +%Y-%m-%d
