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).
Related
I am getting results to a log file that contain a line like this:
date: Sat, 12 Dec 2020 22:33:34 GMT
I want to use only Bash and GNU tools on my Ubuntu Linux box if possible to convert this to my local time "Eastern" or Michigan/Detroit. It should work even on Daylight Saving Time or if past/before Midnight. I want the result stored in a variable in a common format such as 2020-12-01 for December 1, 2020. One variable for the military time, a second for the date would probably be best. I can calculate the "Sat/Sun/Mon/etc" and probably don't need that anyway.
I would expect the "cut" command could separate out the different fields, but how to deal with GMT?
#!/bin/bash
datime="date: Sat, 12 Dec 2020 22:33:34 GMT"
#magic happens
echo dalocaltime is $dalocaltime and dalocaldate is $dalocaldate
results:
dalocaltime is 14:20:33 and dalocaldate is 2020-01-30
This works for me, but I can't explain the ${datime#* } part
datime="date: Sat, 12 Dec 2020 22:33:34 GMT"
dalocaltime=$(date -d "${datime#* }" '+%R')
dalocaldate=$(date -d "${datime#* }" '+%Y-%m-%d')
echo dalocaltime is $dalocaltime and dalocaldate is $dalocaldate
I have a date represented as 1231533845
I understand that is supposed to be Fri Jan 9 15:44:05 EST 2009
And that on OSX it can be decoded with:
date -j -f %s 1231533845
What's the equivalent command on linux?
You will use the date function, but with the option -d
date -d #1231533845
I have an output from a shell script like this:
aaa.sh output
Tue Mar 04 01:00:53 2014
Time drift detected. Please check VKTM trace file for more details.
Tue Mar 04 07:21:52 2014
Time drift detected. Please check VKTM trace file for more details.
Tue Mar 04 13:17:16 2014
Time drift detected. Please check VKTM trace file for more details.
Tue Mar 04 16:56:01 2014
SQL> ALTER DISKGROUP fra ADD DISK '/dev/rhdisk20'
Wed Mar 05 00:03:42 2014
Time drift detected. Please check VKTM trace file for more details.
Wed Mar 05 04:13:39 2014
Time drift detected. Please check VKTM trace file for more details.
Tue Mar 05 05:56:07 2014
GMON querying group 3 at 10 for pid 18, osid 27590856
GMON querying group 3 at 11 for pid 18, osid 27590856
I need to get the part, beginning from today's date:
Wed Mar 05 00:03:42 2014
Time drift detected. Please check VKTM trace file for more details.
Wed Mar 05 04:13:39 2014
Time drift detected. Please check VKTM trace file for more details.
Tue Mar 05 05:56:07 2014
GMON querying group 3 at 10 for pid 18, osid 27590856
GMON querying group 3 at 11 for pid 18, osid 27590856
You can get the date in the correct format like this:
today=$(date +'%a %b %d')
and then search for it like this:
grep "$today" aaa.sh
If there are lines from today without a date, such as your GMON lines, you could add -A to say how many lines after the match you want and use a big number:
grep -A 999999 "$today" aaa.sh
If you are on AIX and there is no -A option, use sed like this:
today=$(date +'%a %b %d')
sed -n "/${today}/,$ p" aaa.sh
Explanation:
That says store today's date in the variable today in the format "Wed Mar 05". Then search, without printing anything (-n) till you find that date, From that point on, till the end of file ($) print all lines (p).
I think I have an easy solution:
Get date to output the date in a format that would match the date in the file (check man date on formatting options). Since we don't want to match the hours/minutes/seconds we have to call date twice: once for the weekday/month/day half and once for the year half on the end of the full date. Between these two halves we match the horus/minutes/seconds with .* regex.
Then do:
aaa.sh | grep -E '`date --only-weekday-month-day`.*`date --only-year`' -A 999999
though I am using answer by NewWorld it can be modified as,
convert output of date similar to your file format
suppose in variable 'D'you get that output
sed '1,/${D}/d' aaa.sh
that will output all lines after match date match.
example: suppose you get D="Wed Mar 05 00:03:42 2014"
output will be as expected.
You can use
tail -n 7 filename
for getting the desired output . It will basically give you the last seven lines of the text file named filename .
For getting solution from today's date you can use :
k=$(date +"%a %b %d")
g=$(grep -nr "$k" in|cut -f1 -d:|head -1)
total=$(wc -l<in)
l=`expr $total - $g + 1
tail -n$l in
Try
sed -n '/Wed Mar 05/,$p' aaa.sh
Here -n means "don't print anything unless specified to".
First appearance of a line that matches the expression /Wed\ Mar\ 05/ till the end of the file, will be printed(p)"
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 :)
I am using Ubuntu 10.04 LTS. In bash I am getting error while converting string to date like this:
date -d '20110327 02:00'
date: invalid date `20110327 02:00'
but these work:
date -d '20110327 03:00'
Sun Mar 27 03:00:00 CEST 2011
date -d '20110326 02:00'
Sat Mar 26 02:00:00 CET 2011
date -d '20110328 02:00'
Mon Mar 28 02:00:00 CEST 2011
Any ideas?
Thanks,
Jan
It's the summer time !
In 2011, we get an additionnal hour on the March 27th at 2:00 it was in fact 3:00.
So 27/03/2011 2:00 is not a valid date :-)
Nope, I'm sure there was a transition to/from Daylight Saving Time (DST) at 2:00 that day, so there wasn't 2:00 :)
Well there is no such local time - there was switch to summer time and the clock went from 1:59 to 3:00.