How to increment a date in a Bash script

Question

I have a Bash script that takes an argument of a date formatted as yyyy-mm-dd.

I convert it to seconds with

startdate="$(date -d"$1" +%s)"; 

What I need to do is iterate eight times, each time incrementing the epoch date by one day and then displaying it in the format mm-dd-yyyy.

Answer 1

Use the date command's ability to add days to existing dates.

The following:

DATE=2013-05-25  for i in {0..8} do    NEXT_DATE=$(date +%m-%d-%Y -d "$DATE + $i day")    echo "$NEXT_DATE" done 

produces:

05-25-2013 05-26-2013 05-27-2013 05-28-2013 05-29-2013 05-30-2013 05-31-2013 06-01-2013 06-02-2013 

Note, this works well in your case but other date formats such as yyyymmdd may need to include "UTC" in the date string (e.g., date -ud "20130515 UTC + 1 day").

Answer 2

startdate=$(date -d"$1" +%s) next=86400 # 86400 is one day  for (( i=startdate; i < startdate + 8*next; i+=next )); do      date -d"@$i" +%d-%m-%Y done