ANSI date numbers starts from January 1st 1601 (day 1).
So how to get the following to work in a bash command in Linux?
I want:
# ANSI / UNIX epoch delta is 134774 days
$ date -ud ‘1601 -01 -01 + 134774 days ’ +%F
1970 -01 -01
But I get
date: invalid date '1601-01-01+134774 days'
To answer my own question as I meanwhile found this site: https://unix.stackexchange.com/questions/7688/date-years-prior-to-1901-are-treated-as-invalid
It's because I'm on a 32-bit machine.
date -ud '1901-12-14 + 24855 days' +%F
will give
1970-01-01
Related
I need to the date three days from today. On CentOS, I can run the below command.
date -I -d "-3 days"
Which outputs
2022-07-02
I need the same inside my Docker container which is running on Alpine Linux 3.14.6v.
When I execute the same command I am getting the error:
date: invalid date '-3 days'
Anyone knows the workaround for this?
From StackExchange.com:
https://unix.stackexchange.com/questions/206540/date-d-command-fails-on-docker-alpine-linux-container
BusyBox/Alpine version of date doesn't support -d options, even if the
help is exatly the same in the Ubuntu version as well as in others
more fat distros.
To work with -d options you just need to add coreutils package
A way to do it could be to use a bit of arithmetic on a timestamp, then translate the timestamp back into a date:
date -d "#$(( $(date +%s) - 3 * 24 * 60 * 60 ))"
Given
docker run --rm alpine:3.14 sh -c 'date;
date -d "#$(( $(date +%s) - 3 * 24 * 60 * 60 ))"'
It would yield:
Tue Jul 5 11:51:31 UTC 2022
Sat Jul 2 11:51:31 UTC 2022
I am trying to convert dates from different timezones with UNIX date (I am on Ubuntu 20.04).
With current date, it works well:
$ date
dim. 12 déc. 2021 11:59:16 CET
$ TZ=Pacific/Tahiti date
dim. 12 déc. 2021 00:59:32 -10
But when I am working with a string, it fails:
$ export testdate="2021/10/28 17:47:26"
$ date -d "$test"
jeu. 28 oct. 2021 17:47:26 CEST
$ TZ=Pacific/Tahiti date -d "$test"
jeu. 28 oct. 2021 17:47:26 -10
as I am expecting:
jeu 28 oct. 2021 05:47:26 -10
I don't understand why I don't get the proper shift. And of course if I try with a date and time where the day should also change, it doesn't work either:
$ export test="2021/10/28 7:47:26"
$ date -d "$test"
jeu. 28 oct. 2021 07:47:26 CEST
$ TZ=Pacific/Tahiti date -d "$test" "+%F %T %Z"
2021-10-28 07:47:26 -10
while I am expecting:
mer 27 oct. 2021 19:47:26 -10
why I don't get the proper shift
test="2021/10/28 17:47:26"
Is a date in unknown timezone. No one knows what timezone it is in, what the daylight is. GNU date tries to "guess" what offset you meant, it generally traverses TZ database for current timezone and just picks the first offset that matches. Also, specifying timezone is not enough to know what daylight it is, you have to be specific.
Also, because of the daylight time you can "go back" in time, it's now known what the offset to UTC is even when you know the timezone.
Also, you don't have to export it - date does not care about test environment variable.
Converting date in a different timezone with date
If the input is in UTC, tell date that.
$ LC_ALL=C TZ=Pacific/Tahiti date -d "2021/10/28 17:47:26 UTC"
Thu Oct 28 07:47:26 -10 2021
If the input is with any other offset, tell date that.
$ LC_ALL=C TZ=Pacific/Tahiti date -d "2021/10/28 17:47:26 CEST"
Thu Oct 28 05:47:26 -10 2021
Te parsing of GNU date of input format is generally a mystery. The documentation lists 2004-02-29 16:21:42 format as an example input, so I recommend that format. If you want to be exact, I recommend strptime from dateutils (or a real programming language).
One simple way is to convert time first to epoch time :
test="2021/10/28 17:47:26"
TZ=Pacific/Tahiti date -d #$(date -d "$test" +%s)
date -d "$test" +%s converts local time to epoch time.
TZ=Pacific/Tahiti date -d #$(date -d "$test" +%s) prints Tahiti time from epoch time.
I'm not fully confident, but i get the impression you're looking the following syntax:
LC_TIME="es_ES.UTF8" TZ="America/New_York" date --date='TZ="Europe/Amsterdam" 2021/10/28 17:47:26' "+%A %F %T %B"
That takes a predefined datetime (interpreted as being local to Amsterdam), adjusts the datetime (based on the time difference) to New York-time at that same moment; Then it prints that result with Spanish names for the months/weekdays (provided that language' locale is present on your system).
How to set a cron to execute every 10 days starting from 16th January? Would this suffice?
30 7 16-15/10 * * command >/dev/null
The above starts at 7.30 AM, 16th of every month and ends on next month 15th and repeats every 10 days. I don't think what I have above is correct. Can anyone tell me how to set up the cron so that month ends are taken into account and every 10 days the command is executed starting from 16th January this year 2016?.
As William suggested, cron can't handle this complexity by itself. However, you can run a cron job more frequently, and use something else for the logic. For example;
30 7 16-31 1 * date '+\%j' | grep -q '0$' && yourcommand
30 7 * 2-12 * date '+\%j' | grep -q '0$' && yourcommand
This date format string prints the day of the year, from 001 to 365. The grep -q will do a pattern match, NOT print the results, but return a success of a failure on the basis of what it finds. Every 10 days, the day of the year ends in a zero. On those days, yourcommand gets run.
This has a problem with the year roll-over. A more complex alternative might be to do a similar grep on a product of date '+%s' (the epoch second), but you'll need to do math to turn seconds into days for analysis by grep. This might work (you should test):
SHELL=/bin/bash
30 7 * * * echo $(( $(date '+%s') / 86400 )) | grep '0$' && yourcommand
(Add your Jan 16th logic too, of course.)
This relies on the fact that shell arithmetic can only handle integers. The shell simply truncates rather than rounding.
UPDATE
In a comment on another answer, you clarified your requirements:
The command should start executing on January 16th, and continue like on January 26th, February 5th, February 15th and so on – jai
For this, the epoch-second approach is probably the right direction.
% date -v1m -v16d -v7H -v30M -v0S '+%s'
1452947400
(I'm in FreeBSD, hence these arguments to date.)
SHELL=/bin/bash
30 7 * * * [[ $(( ($(date '+\%s') - 1452947400) \% 864000 )) == 0 ]] && yourcommand
This expression subtracts the epoch second of 7:30AM Jan 16 (my timezone) from the current time, and tests whether the resultant difference is divisible by 10 days. If it is, the expression evaluates true and yourcommand is run. Note that $(( 0 % $x )) evaluates to 0 for any value of $x.
This may be prone to error if cron is particularly busy and can't get to your job in the one second where the math works out.
If you want to make this any more complex (and perhaps even if it's this complex), I recommend you move the logic into a separate shell script to handle the date comparison math. Especially if you plan to add a fudge factor to allow for jobs to miss their 1-second window .. that would likely be multiple lines of script, which is awkward to maintain in a single cronjob entry.
Observation: the math capabilities of cron are next to non-existent. The math capabilities of the Unix tools are endless.
Conclusion: move the problem from the cron domain to the shell domain.
Solution: run this each day with 30 7 * * * /path/to/script in the crontab:
#!/bin/sh
PATH=$(/usr/bin/getconf PATH)
if test $(($(date +%j) % 10)) = 6; then
your_command
fi
This tests whether the day-of-year modulo 10 is 6, like it is for January 16 (and January 6th is already in the past...).
Thinking outside the box:
Fix your requirement. Convince whoever came up with that funny 10 day cycle to accept a 7 day cycle. So much easier for cron. This is following the KISS principle.
0 30 7 1/10 * ? * command >/dev/null
Output for the above express is,
Saturday, January 16, 2016 7:30 AM
1. Thursday, January 21, 2016 7:30 AM
2. Sunday, January 31, 2016 7:30 AM
3. Monday, February 1, 2016 7:30 AM
4. Thursday, February 11, 2016 7:30 AM
5. Sunday, February 21, 2016 7:30 AM
Output for your expression
i.e 30 7 16-15/10 * * command >/dev/null
2016-01-15 07:30:00
2016-02-15 07:30:00
2016-03-15 07:30:00
2016-04-15 07:30:00
2016-05-15 07:30:00
2016-06-15 07:30:00
2016-07-15 07:30:00
2016-08-15 07:30:00
2016-09-15 07:30:00
2016-10-15 07:30:00
The closest syntax would like this:
30 7 1-30/10 * *
30 7 1-31/10 * *
30 7 1-28/10 * *
30 7 1-29/10 * *
You can test the cron expression here http://cron.schlitt.info/
In linux bash when I enter date -d "1986-01-01" it shows error
date: invalid date "1986-01-01"
when date -d "1986-01-02" it works
when date -d "1987-01-01" it also works
Why date -d "1986-01-01" shows error in Linux Bash shell.
I am using Fedora 16
Nepal changed its timezone at the beginning of 1986. The following table is copied from the tzdata package:
# Zone NAME GMTOFF RULES FORMAT [UNTIL]
Zone Asia/Kathmandu 5:41:16 - LMT 1920
5:30 - IST 1986
5:45 - NPT # Nepal Time
That means that on Jan 1 1986 the time from 00:00:00 to 00:14:59 is not valid. The following two commands show, that the first day of 1986 started with 00:15:00:
$ TZ=Asia/Kathmandu date -d '1985-12-31 23:59:59' '+%s'
504901799
$ TZ=Asia/Kathmandu date -d '1986-01-01 00:15:00' '+%s'
504901800
So the error message of date is correct. The date is invalid in this timezone. I am not sure what you are doing with the result of this command. However, you can try to use UTC because all dates are valid and unambiguous in UTC:
$ TZ=UTC date -d '1986-01-01'
Wed Jan 1 00:00:00 UTC 1986
I think you are using alphabet "O" in upper case instead of number "0" in the command :)
The command
mydate=$(date -d "90 days 19850101" +%Y%m%d%H%M%S)
yields 19850401000000. But:
mydate=$(date -d "90 days 19830101" +%Y%m%d%H%M%S)
yields 19830401010000.
How is it possible that in year 1983 one hour is added on 1 April (which is a result I don't want), while for the year 1985 the answer is correct?