I'm working on a script to automate the download of information from a customer in the period from 00:00 to 23:59 on the same day. To make the correct treatment of the first day of daylight saving time (10/15/2017, in my timezone - BRST), I need to check the timezone of the previous day. However, when I will subtract one day from the first valid time,
date --date="20171015 01:00 -1 day" +%Y-%m-%d
the result is the next day 2017-10-16, not the previous day 2017-10-14. Could anyone help me understand what I might be doing wrong and how should I do this operation in the right way?
I don't really trust GNU date's arithmetic. Instead, I would convert your starting time to seconds since the UNIX epoch, subtract 86400 seconds, and convert the result back to a day. Those conversion routines take daylight saving time into account.
$ TZ=BRST date +%s --date "20171015 01:00"
1508029200
$ TZ=BRST date +%F-%T --date #$((1508029200 - 86400))
2017-10-14-01:00:00
It appears the "BRST" timezone is not very useful: stick to Olson timezone names
$ TZ=BRST date --date="20171015 00:30" "+%F %T"
2017-10-15 00:30:00
$ TZ=America/Sao_Paulo date --date="20171015 00:30" "+%F %T"
date: invalid date ‘20171015 00:30’
Related
I have a file "file_XYZ_18548".
Here the name's ending 18548 is a timestamp in days, and it is changing day by day, like "file_XYZ_18550".
I would like to get this date via variable but I couldn't find a date command to get the timestamp in days.
I can get the date result but I can't get the timestamp in reverse.
timeinday=18550
timestp=timeinday*86400
datetm=$(echo $timestp | gawk '{print(strftime("%Y-%m-%d %H:%M:%S", $0))}')
echo $datetm
2020-10-14 10:09:10
How can I get this with date command in bash scripting? Is there any way of this via awk/gawk etc..?
Given that 2020-10-14 - 18548 days corresponds to 1970-01-03, it is reasonable to guess that the 'epoch' for the day count is 1970-01-01. It is likely that day 1 was 1970-01-01, so day zero was 1969-12-31.
Converting day count to date
You can use the GNU date command like this:
daycount=18548
date -u -d "#$(( ($daycount-1) * 86400 ))" +"%Y-%m-%d %H:%M:%S"
That yields the result 2020-10-12 00:00:00. You can drop the time component of the format if you wish (you probably do; midnight isn't very exciting when it is always midnight). You can calibrate the -1 to resolve exactly what day should correspond to 18548. Just in case it isn't obvious, there are 86,400 seconds in a day (24 hours • 60 minutes per hour • 60 seconds per minute).
The $(( … )) notation is Bash's Arithmetic Expansion notation.
Converting current time to day count
If you want to convert the current time to the day offset, then you can use the %s specifier to get the seconds since 1970-01-01 00:00:00Z and divide by 86400 to get the number of days:
echo $(( $(date +'%s') / 86400 ))
which (at 2020-10-14 23:30 -06:00, aka 1602739800 seconds since the Unix Epoch) yields the result:
18550
Again, if need be, you can adjust the value of the division to account for when day 1 was in this scheme. Shell arithmetic in Bash is integer arithmetic, which is exactly what is wanted. You might need to use -u to get UTC as the time zone (as I did earlier), and so on.
I'm trying to calculate the seconds between a date from now, however, I always get extra one day when adding the seconds from now.
echo $(($(date -ud "2020-03-15 19:13" +'%s') - $(date +'%s')))
As of posting, the result is 1744204
Using this website to check, I gets 16 March, not 15 March as expected. Any idea why?
verify the result of your date commands and verify the timezone of both.
The first result of the command shows the timestamp in UTC, and the second one shows the timestamp using the timezone of the system.
Here is the difference:
$ date -ud "2020-03-15 19:13" +'%s'
1584299580
With UTC-3 in my system:
$ date -d "2020-03-15 19:13" +'%s'
1584310380
I hope that help you.
When I issue bash command:
date --date="2018-03-03 12:16:13 -1hour" "+%Y:%m:%d %H:%M:%S"
I expect the result would be:
2018:03:03 11:16:13
but instead, I get:
2018:03:03 15:16:13
I wonder if this has to make with time zones, and how to avoid this behaviour.
I can reproduce this. My timezone is America/New_York
$ date --date="2018-03-03 12:16:13 - 1 hour" "+%Y:%m:%d %H:%M:%S"
2018:03:03 09:16:13
$ env TZ='Europe/Belgrade' date --date="2018-03-03 12:16:13 - 1 hour" "+%Y:%m:%d %H:%M:%S"
2018:03:03 15:16:13
The parser appears to be taking the -1 as the timezone GMT+01:00, then converting that to your local timezone.
If we rearrange the phrases to avoid the timezone parsing ambiguity, we can get your desired result:
$ date --date="- 1 hour 2018-03-03 12:16:13" "+%Y:%m:%d %H:%M:%S"
2018:03:03 11:16:13
From info coreutils 'date invocation'
When a relative item causes the resulting date to cross a boundary where the clocks were adjusted, typically for daylight saving time,
the resulting date and time are adjusted accordingly.
The fuzz in units can cause problems with relative items. For
example, '2003-07-31 -1 month' might evaluate to 2003-07-01, because
2003-06-31 is an invalid date. To determine the previous month more
reliably, you can ask for the month before the 15th of the current
month. For example:
$ date -R
Thu, 31 Jul 2003 13:02:39 -0700
$ date --date='-1 month' +'Last month was %B?'
Last month was July?
$ date --date="$(date +%Y-%m-15) -1 month" +'Last month was %B!'
Last month was June!
Also, take care when manipulating dates around clock changes such
as daylight saving leaps. In a few cases these have added or
subtracted as much as 24 hours from the clock, so it is often wise to
adopt universal time by setting the 'TZ' environment variable to
'UTC0' before embarking on calendrical calculations.
One can avoid that by putting -1 hour before the string,
$ date --date='-1 hour 2018-03-03 12:16:13' "+%Y:%m:%d %H:%M:%S"
2018:03:03 11:16:13
I am trying to do basic time math in Bash having following timestamps from OpenDj:
createtimestamp: 20161123165725Z
I don't know what time zone the system will be set with, it can be with "Zulu" timezone (UTC) as above, or might be with some other timezone. The above timestamp is for the users on LDAP servers. What I need to do is to compare the timestamp of the user with a certain timestamp.
Question: how do I convert the timestamp with zone (Z) above to local timestamp or epoch time in Bash?
Maybe Dateutils will help
http://www.fresse.org/dateutils/
dateconv -i '%s' -f '%A, %d %b %Y' 1234567890
Friday, 13 Feb 2009
dateutils tools default to UTC (add -z your/timezone if needed).
How to get in bash diff between utc time and localtime in seconds ?
%s isn't trivially part of the answer without some time-related antics, epoch seconds is explicitly UTC (or maybe GMT if you're old-school), so setting TZ won't affect it.
Similar to twalberg's suggestion:
IFS=":" read hh mm < <(date +%:z)
echo $(($hh*3600+$mm*60))
You can test that this is doing what you want by setting TZ for the date command:
IFS=":" read hh mm < <(TZ=Australia/Sydney date %:z) # answer is 39600
IFS=":" read hh mm < <(TZ=US/Eastern date +%:z) # answer is -18000
(This isn't strictly a bash answer, since it requires GNU date or equivalent, 5.90 and later support %:z and %::z)
You'll probably have to do a little work to get it in seconds, but date +%z will report your configured timezone offset as [+-]HHMM. Parse the output of that and do the appropriate math to figure it out in seconds.
This will only work in places with whole hours and not minutes.
echo $(($( date +%z)*36))