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For example, I have date: 4 August 1993 and I want to add 348 days to it, how can I do it in bash?

inothemo
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3 Answers3

35

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
David Webb
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19

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

Tom Söderlund
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    Been 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 G Jan 15 '19 at 21:11
  • Nice! There's a whole slew of formatting options you can use with the Bash `date` command at https://www.tutorialkart.com/bash-shell-scripting/bash-date-format-options-examples/ – AFK Sep 18 '21 at 03:56
  • You can't believe how many incorrect answers I've stumbled upon until this one. Thanks! – Matthew Oct 06 '22 at 18:03
1

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
Bohdan
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