How to generate a sequence of dates given starting and ending dates using AWK of BASH scripts?

Question

I have a data set with the following format

The first and second fields denote the dates (M/D/YYYY) of starting and ending of a study.

How one expand the data into the desired output format, taking into account the leap years using AWK or BASH scripts?

Your help is very much appreciated.

Input

  7/2/2009   7/7/2009
  2/28/1996  3/3/1996
  12/30/2001 1/4/2002

Desired Output

  7/7/2009
  7/6/2009
  7/5/2009
  7/4/2009
  7/3/2009
  7/2/2009
  3/3/1996
  3/2/1996
  3/1/1996
  2/29/1996
  2/28/1996
  1/4/2002
  1/3/2002
  1/2/2002
  1/1/2002
  12/31/2001
  12/30/2001

Answer 1

It can be done nicely with bash alone:

for i in `seq 1 5`;
do
  date -d "2017-12-01 $i days" +%Y-%m-%d;
done;

or with pipes:

seq 1 5 | xargs -I {} date -d "2017-12-01 {} days" +%Y-%m-%d

Answer 2

If you have gawk:

#!/usr/bin/gawk -f
{
    split($1,s,"/")
    split($2,e,"/")
    st=mktime(s[3] " " s[1] " " s[2] " 0 0 0")
    et=mktime(e[3] " " e[1] " " e[2] " 0 0 0")
    for (i=et;i>=st;i-=60*60*24) print strftime("%m/%d/%Y",i)
}

Demonstration:

./daterange.awk inputfile

Output:

07/07/2009
07/06/2009
07/05/2009
07/04/2009
07/03/2009
07/02/2009
03/03/1996
03/02/1996
03/01/1996
02/29/1996
02/28/1996
01/04/2002
01/03/2002
01/02/2002
01/01/2002
12/31/2001
12/30/2001

Edit:

The script above suffers from a naive assumption about the length of days. It's a minor nit, but it could produce unexpected results under some circumstances. At least one other answer here also has that problem. Presumably, the date command with subtracting (or adding) a number of days doesn't have this issue.

Some answers require you to know the number of days in advance.

Here's another method which hopefully addresses those concerns:

while read -r d1 d2
do
    t1=$(date -d "$d1 12:00 PM" +%s)
    t2=$(date -d "$d2 12:00 PM" +%s)
    if ((t2 > t1)) # swap times/dates if needed
    then
        temp_t=$t1; temp_d=$d1
        t1=$t2;     d1=$d2
        t2=$temp_t; d2=$temp_d
    fi
    t3=$t1
    days=0
    while ((t3 > t2))
    do
        read -r -u 3 d3 t3 3<<< "$(date -d "$d1 12:00 PM - $days days" '+%m/%d/%Y %s')"
        ((++days))
        echo "$d3"
    done
done < inputfile

Answer 3

You can do this in the shell without awk, assuming you have GNU date (which is needed for the date -d @nnn form, and possibly the ability to strip leading zeros on single digit days and months):

while read start end ; do
    for d in $(seq $(date +%s -d $end) -86400 $(date +%s -d $start)) ; do
        date +%-m/%-d/%Y -d @$d
    done
done

If you are in a locale that does daylight savings, then this can get messed up if requesting a date sequence where a daylight saving switch occurs in between. Use -u to force to UTC, which also strictly observes 86400 seconds per day. Like this:

while read start end ; do
    for d in $(seq $(date -u +%s -d $end) -86400 $(date -u +%s -d $start)) ; do
        date -u +%-m/%-d/%Y -d @$d
    done
done

Just feed this your input on stdin.

The output for your data is:

7/7/2009
7/6/2009
7/5/2009
7/4/2009
7/3/2009
7/2/2009
3/3/1996
3/2/1996
3/1/1996
2/29/1996
2/28/1996
1/4/2002
1/3/2002
1/2/2002
1/1/2002
12/31/2001
12/30/2001