How to print dates between two dates in format %Y%m%d in shell script?

Question

I have two arguments as inputs: startdate=20160512 and enddate=20160514.

I want to be able to generate the days between those two dates in my bash script, not including the startdate, but including the enddate:

20160513 20160514 I am using linux machine. How do I accomplish this? Thanks.

Answer 1

Using GNU date:

$ d=; n=0; until [ "$d" = "$enddate" ]; do ((n++)); d=$(date -d "$startdate + $n days" +%Y%m%d); echo $d; done
20160513
20160514

Or, spread over multiple lines:

startdate=20160512
enddate=20160514
d=
n=0
until [ "$d" = "$enddate" ]
do  
    ((n++))
    d=$(date -d "$startdate + $n days" +%Y%m%d)
    echo $d
done

How it works

• d=; n=0

  Initialize variables.

• until [ "$d" = "$enddate" ]; do

  Start a loop that ends on enddate.

• ((n++))

  Increment the day counter.

• d=$(date -d "$startdate + $n days" +%Y%m%d)

  Compute the date for n days after startdate.

• echo $d

  Display the date.

• done

  Signal the end of the loop.

Answer 2

Another option is to use dateseq from dateutils (http://www.fresse.org/dateutils/#dateseq). -i changes the input format and -f changes the output format.

$ dateseq -i%Y%m%d -f%Y%m%d 20160512 20160514
20160512
20160513
20160514
$ dateseq 2016-05-12 2016-05-14
2016-05-12
2016-05-13
2016-05-14

Answer 3

This should work on OSX, make sure your startdate is lesser than enddate, other wise try with epoch.

startdate=20160512
enddate=20160514

loop_date=$startdate

let j=0
while [ "$loop_date" -ne "$enddate" ]; do
        loop_date=`date   -j -v+${j}d  -f "%Y%m%d" "$startdate" +"%Y%m%d"`
        echo $loop_date
        let j=j+1
done