Linux command to check wherever this day is a DST change day and which direction?

Question

Im currently designing a solution to synchronize a Clock on a Equipment. Each time the Equipment is turned on after a poweroff, its Clock is exactly "01:01". To sync this Equipment, I have a NEXA remote switch and a Telldus duo. The Telldus duo is connected to a server which is a NTP server with GPS and PPS source.

Now to the problem: Currently, the Clock synchronizes by simply turning off Power scheduled at 01:00, and then turning on Power 01:01, causing the Clock on the Equipment to Always go correct.

But the problem is on DST change Days. If there is a DST change day that says the Clock should be moved FORWARD 1 hour at 02:00 to 03:00, then the Clock on the Equipment will be behind by 1 hour in one full day. If there is a DST change day that says the Clock should be moved BACK 1 hour, then the equpment will be running 1 hour too fast.

Thus, on the DST FORWARD Days, I need to Schedule the Equipment to turn off at 00:00 and turn on at 00:01. Then the Equipment will be 01:01 when Clock is 00:01. When Clock is 02:00 and is changed to 03:00, then Equipment will already be at 03:00, thus Equipment will only have incorrect Clock for 2 hour.

And on the DST BACK Days, I need to Schedule the Equipment to turn off at 02:00 and turn on at 02:01. Thus the Equipment will be 02:00 when Clock is 03:00 and being DST'ed back to 02:00, thus in this case Equipment's Clock will only be incorrect for 1 hour.

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I have the full capatibility to write a script that will accomplish this, GIVEN THAT:

I know a command, that can tell me, if TODAY is the day that a DST change will occur, and in which direction (02:00 to 03:00, or 03:00 to 02:00) it will occur.

This means I need either some dedicated DST command, OR a possibility to "look in the future", like "What is the Clock when 4 hours have elapsed from now, accounting for DST?". (Then I could check, if the difference between now and returned clock is 3 hours, then its a DST-BACK day, 4 hours, then its a NO-DST day and 5 hours then its a DST-FORWARD day)

Anyone that have a idea of such command?

Answer 1

Try:

$ zdump -v Europe/London  -c $(date '+%Y'),$(date -d '+1 year' '+%Y') | awk -v y=$(date '+%Y') '$6==y && $15~1 {print $4, $3, $6}'
30 Mar 2014
26 Oct 2014

Which prints the dates at which DST changes. Change the Europe/London to your timezone.

Or, slightly longer:

$ zdump -v Europe/London  -c $(date '+%Y'),$(date -d '+1 year' '+%Y') | awk -v y=$(date '+%Y') '
                                     ($6==y && dst=="") {dst=$15}
                                     ($6==y && dst==$15) {ftime=$12; dst=$15;}
                                     ($6==y && dst!=$15) {print ftime, "to", $12, "on", $4, $3, $6, $15; dst=$15}'
00:59:59 to 02:00:00 on 30 Mar 2014 isdst=1
01:59:59 to 01:00:00 on 26 Oct 2014 isdst=0

Which prints the time changes too. The last column shows whether you are entering DST (1 means entering DST).

Answer 2

If anyone has come here via Google looking for a simple daylight savings detection command for the linux terminal, this suffices. This is not complex enough for OP's question however.

if perl -e 'exit ((localtime)[8])' ; then
    echo no DST
else
    echo DST
fi

*the 8th element in the output of localtime in perl is $isdst http://perldoc.perl.org/functions/localtime.html

Answer 3

I would just ask the system how it would print the time in exactly 24 hours:

now=$(date +%H)
then=$(date -d '24 hours' +%H)
dst_change=$(( then - now ))

Caution, don't run this on the hour sharp...