how to create a data file with a column of date from day1 to dayN?

Question

I was wondering if I could generate a hourly or daily time step date in a column starting from some year in the past to some day in the near past or today. to be more clear, I want to create one column of data from 2000-10-10 100 to 2012-12-31 2400. The output file will look like

Date 

2000-01-01 100
2000-01-01 200
2000-01-01 300
.
.
.
.
.
2012-12-31 2400

Answer 1

#!/bin/bash

day=2000-01-01
end=2012-12-31

echo Date > output.file

until [[ $day > $end ]]; do
    printf "$day %d\n" $(seq 100 100 2400)
    day=$(date -d "$day + 1 day" +"%Y-%m-%d")
done >> output.file

Answer 2

With ksh93:

d1=$(printf '%(%s)T\n' "2000-01-01 01:00:00")
d2=$(printf '%(%s)T\n' "2013-01-01 00:00:00")
for ((d=d1;d<d2;d+=3600)); do
  printf '%(%F %-H%M)T\n' "#$d"
done

With zsh:

zmodload zsh/datetime
strftime -rs d1 '%Y%m%d%H%M' 201201010100
strftime -rs d2 '%Y%m%d%H%M' 201301010000
for ((d=d1;d<d2;d+=3600)) strftime '%F %-H%M' $d

With perl:

perl -MPOSIX -le '
  $d1=mktime 0,0,1,1,0,100;$d2=mktime 0,0,0,1,0,113;
  for ($d=$d1; $d<$d2; $d+=3600) {
    print strftime "%F %-H%M", localtime $d}'

With GNU awk:

awk 'BEGIN {d1=mktime("2000 01 01 01 00 00")
            d2=mktime("2013 01 01 00 00 00")
            for (d=d1;d<d2;d+=3600)
              print strftime("%F %-H%M",d)}'

Note that with all the above around DST changing time some hours will be skipped or output twice if in a timezone with DST as a clock would.

We're counting from 000 to 2300 instead of 100 to 2400.

Answer 3

Since the question has gotten more understandable, the following little script could do something similar.

#!/bin/bash

i=1
c=100

while ((c>i))
    do
    echo "$(date +%F)" $c >> thisfile
    sleep 5
    echo "$(date +%F)" $((c+=100))
    done

I hope it's self explanatory

I've used a sleep 5 but to get exactly this line each hour generated you can use something like sleep $((60*60))