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Linux local time versus UTC in seconds

Could you please explain what is wrong with this 2 syntax for getting the elapsed time in seconds?

If I run both of these commands at the same time:

date -u +%s   # - for UTC - returns e.g. 1303430843. 

date +%s     # - for local time - mine is set to pacific time
          # - also returns e.g. 1303430843

I get exactly the same number (e.g. 1303430843) seconds, no difference, but I would expect 25200 seconds - a 7 hour difference, as if I ran:

date -u "+%Y-%m-%d %H:%M:%S"    # returns:  2011-04-22 00:01:14

date    "+%Y-%m-%d %H:%M:%S"    # returns:  2011-04-21 17:01:14 - 7 hours difference

What's the trick here?

thank

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4 answers


From the man page:

% s seconds since 1970-01-01 00:00:00 UTC



so it %s

returns the time in seconds since a specific point in time specified in UTC, which means it doesn't depend on time zones at all.

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Don't catch. The clock is set and starts in UTC format. The date command takes local time zone and day savings into account. This means that you can display the date / time for any time zone in the world.



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This is because GMT uses UTC (they are mostly used interchangeably) and PDT is currently 7 hours behind .

To get the current UTC time in seconds:

date -u +%s

To get the current PDT time in seconds:

TZ=":US/Pacific" date +%s

It's smart enough to pick up DST so you don't have to worry about it.

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%s     seconds since 1970-01-01 00:00:00 UTC

This does not change with the time zone. The calculation is the same.

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