Bash - number of hours in given day

Question

Using Bash shell on Linux, and given a datetime, how can I determine how many hours there are on that particular day?

The datetime pertains to some time zone with daylight saving, eg MET.

Answer 1

In order to completely account for all scenarios, you need to consider a few things:

• Not every local day has a midnight, and the date command will fail if you pass a date on one of these days, unless you also pass a time and an offset from UTC. This primarily occurs on the spring-forward transition days. For example:

   $ TZ=America/Sao_Paulo date -d '2016-10-16' date: invalid date '2016-10-16'

• Not every DST transition is 1 hour. America/Lord_Howe switches by 30 minutes. Bash only performs integer division, so you have to use one of these techniques if you want decimals.

Here is a function that accounts for these:

seconds_in_day() {
  # Copy input date to local variable
  date=$1

  # Start with the offset at noon on the given date.
  # Noon will almost always exist (except Samoa on 2011-12-30)
  offset1=$(date -d "$date 12:00" +%z)

  # Next get the offset for midnight.  If it doesn't exist, the time will jump back to 23:00 and we'll get a different offset.
  offset1=$(date -d "$date 00:00 $offset1" +%z)

  # Next get the offset for the next day at midnight.  Again, if it doesn't exist, it will jump back an hour.
  offset2=$(date -d "$date 00:00 $offset1 + 1 day" +%z)

  # Get the unix timestamps for both the current date and the next one, at midnight with their respective offsets.
  unixtime1=$(date -d "$date 00:00 $offset1" +%s)
  unixtime2=$(date -d "$date 00:00 $offset2 + 1 day" +%s)

  # Calculate the difference in seconds and hours.  Use awk for decimal math.
  seconds=$((unixtime2 - unixtime1))
  hours=$(awk -v seconds=$seconds 'BEGIN { print seconds / 3600 }')

  # Print the output
  echo "$date had $seconds secs in $TZ, or $hours hours."
}

Examples:

$ TZ=America/Los_Angeles seconds_in_day 2016-03-12
2016-03-12 had 86400 secs in America/Los_Angeles, or 24 hours.
$ TZ=America/Los_Angeles seconds_in_day 2016-03-13
2016-03-13 had 82800 secs in America/Los_Angeles, or 23 hours.
$ TZ=America/Los_Angeles seconds_in_day 2016-03-14
2016-03-14 had 86400 secs in America/Los_Angeles, or 24 hours.

$ TZ=America/Los_Angeles seconds_in_day 2016-11-05
2016-11-05 had 86400 secs in America/Los_Angeles, or 24 hours.
$ TZ=America/Los_Angeles seconds_in_day 2016-11-06
2016-11-06 had 90000 secs in America/Los_Angeles, or 25 hours.
$ TZ=America/Los_Angeles seconds_in_day 2016-11-07
2016-11-07 had 86400 secs in America/Los_Angeles, or 24 hours.

$ TZ=America/Sao_Paulo seconds_in_day 2016-02-19
2016-02-19 had 86400 secs in America/Sao_Paulo, or 24 hours.
$ TZ=America/Sao_Paulo seconds_in_day 2016-02-20
2016-02-20 had 90000 secs in America/Sao_Paulo, or 25 hours.
$ TZ=America/Sao_Paulo seconds_in_day 2016-02-21
2016-02-21 had 86400 secs in America/Sao_Paulo, or 24 hours.

$ TZ=America/Sao_Paulo seconds_in_day 2016-10-15
2016-10-15 had 86400 secs in America/Sao_Paulo, or 24 hours.
$ TZ=America/Sao_Paulo seconds_in_day 2016-10-16
2016-10-16 had 82800 secs in America/Sao_Paulo, or 23 hours.
$ TZ=America/Sao_Paulo seconds_in_day 2016-10-17
2016-10-17 had 86400 secs in America/Sao_Paulo, or 24 hours.

$ TZ=Australia/Lord_Howe seconds_in_day 2016-04-02
2016-04-02 had 86400 secs in Australia/Lord_Howe, or 24 hours.
$ TZ=Australia/Lord_Howe seconds_in_day 2016-04-03
2016-04-03 had 88200 secs in Australia/Lord_Howe, or 24.5 hours.
$ TZ=Australia/Lord_Howe seconds_in_day 2016-04-04
2016-04-04 had 86400 secs in Australia/Lord_Howe, or 24 hours.

$ TZ=Australia/Lord_Howe seconds_in_day 2016-10-01
2016-10-01 had 86400 secs in Australia/Lord_Howe, or 24 hours.
$ TZ=Australia/Lord_Howe seconds_in_day 2016-10-02
2016-10-02 had 84600 secs in Australia/Lord_Howe, or 23.5 hours.
$ TZ=Australia/Lord_Howe seconds_in_day 2016-10-03
2016-10-03 had 86400 secs in Australia/Lord_Howe, or 24 hours.

Answer 2

Oct 30 was the last summer time change here in Britain. I can get 25 hours for that day from the shell in this way:

t1=$(TZ='Europe/London' date --date='20161030' +%s)
t2=$(TZ='Europe/London' date --date='20161031' +%s)
echo $((($t2 - $t1) / 3600))

I am not totally sure this will work in every bash shell and may need to be adjusted a little bit.