Subtract two dates using Shell Scripting

Question

How can I subtract two dates in epoch format using Shell Scripting. I want the output in Months, Days, Hours format. Also It should work even for more than 12 months ( as I came across few which were resetting to 0 months if more than 12)

didn't it depends on starting date ? 29 days from Jan,31 to Mar,1st (and one month), 29 days from July,1 to July,30 and zero month. — Archemar, Oct 12 '15 at 14:24
https://github.com/hroptatyr/dateutils is the best bet so far. — Deer Hunter, Oct 12 '15 at 14:42

score 5 · Accepted Answer · answered Oct 12 '15 at 14:44

5

Try something like this:

#!/bin/bash

d1=`date -d 20140929 +%s`
d2=`date -d 20001115 +%s`

date --date=@$(($d1 - $d2)) +'%m months, %d days, %H hours'

Output:

11 months, 15 days, 02 hours

answered Oct 12 '15 at 14:44

A.P.

1,416
9
11

This resets months count each year – Dani_l Oct 12 '15 at 15:29
Yeah, this doesn't work. Here's what I get: d1="03/26/2022" d2=$(date +%m/%d/%Y) date --date=@$(($d1 - $d2)) +%m-%d-%H 12-31-19 – Jay Imerman Dec 07 '21 at 13:52

score 0 · Answer 2 · answered Oct 12 '15 at 15:13

0

#!/bin/bash

d=(60sec 60min 24hours 30days 12month 1000year)
i=0

while [ $1 -ge ${d[i]%%[a-z]*} ]
do
    set -- $(($1/${d[i]%%[a-z]*})) $(($1%${d[i]%%[a-z]*})) ${d[i]##*[0-9]} ${*:2}
    ((i++))
done 
echo $1 ${d[i]##*[0-9]} ${*:2}

answered Oct 12 '15 at 15:13

Costas

14,916

Subtract two dates using Shell Scripting

2 Answers2