1

On OpenBSD 6.3 I can see:

foo# date
Sun Apr  1 17:00:45 CEST 2018
foo#
foo# date -d "Sun Apr  1 17:00:45 CEST 2018" +%s
1522594858
foo# date
Sun Apr  1 17:00:59 CEST 2018
foo#

that "date" doesn't supports the usual conversion.

The https://www.epochconverter.com/ says that the 1522594858 is:

Your time zone: Sunday, April 1, 2018 5:00:58 PM GMT+02:00 DST

So the date command I issued only returned the unix timestamp of the CURRENT time, not the time that I have given to convert: "Sun Apr 1 17:00:45 CEST 2018".

The Question: how to convert a given date to unix timestamp on OpenBSD? If not with the "date" command, what other shell tools could do it?

First I started to read: https://man.openbsd.org/date - but didn't find anything about converting to unix timestamp. Then I tried to google for many examples, but didn't helped.

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    The man page also mentions a -j flag. That seems like it could be useful. But it seems like the input format must be YYmmddHHMMSS (unlike FreeBSD, there is no -f flag) – Mikel Apr 1 '18 at 15:36
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The OpenBSD date utility is used to

  1. set the date and/or time on the system, or
  2. display the current date and/or time, or
  3. parse the time and date on the format [[[[[[cc]yy]mm]dd]HH]MM[.SS]] without setting the system's time, using the -j option.

It is not date conversion tool that is as flexible as GNU date in its parsing capabilities. If you want GNU date time/date parsing capabilities, install the coreutils port/package and use gdate instead.

Also note that the -d option for OpenBSD's date is used to set the system's value for daylight saving.

0

A smaller executable is busybox. It may do date:

$ busybox date -D '%a %b %-d %T %Z %Y' -d 'Sun Apr  1 17:00:45 CEST 2018' +%s
1522616445

The -D describes the format used to parse the string in -d.
Mind that it does not parse the CEST value, it assumes TZ=UTC.
That is not a problem for a seconds timestamp as it is always at UTC+0.

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