I am trying to subtract 5 minutes from date but its giving unexpected output.
$ date -j -f "%Y/%m/%d %H:%M:%s" -v "-5M" "2021/03/01 09:11:14"
Thu Jan 1 05:25:14 IST 1970
Please suggest the correction.
Converting my comment to answer so that solution is easy to find for future visitors.
This date command should work on BSD date:
date -j -f "%Y/%m/%d %H:%M:%S" -v "-5M" "2021/03/01 09:11:14"
Issue in your command was use of .%s instead of .%S for the second component.
Related
I'm trying to calculate the number of seconds since the Epoch, using date on MacOS BSD.
I can get a one year ago date string:
$ date -v -1y
Tue Apr 21 10:44:47 EST 2020
...but I can't figure out how to convert it into seconds since Epoch. Any suggestions?
Add +%s to tell it to print the datetime as seconds since the epoch:
date -v -1y +%s
The + is a date option to set the output format, and %s is strftime format for "seconds since epoch".
Portability note: while the +%s part is pretty standard and portable (though the %s format is not actually required by POSIX), the -v -1y part is wildly nonportable. With GNU date (e.g. on most Linuxes), you'd use something like --date='1 year ago' instead. On NetBSD, -d '1 year ago' works. Check your local man page to see what your system supports.
What I intent to get is
$ xxx 2019-10-11 <= insert your command
1570752000
The output is timestamp in Oct 11 00:00:00 UTC 2019. I find a good way to do this in gnu, but not in bsd
This should work:
date -j -f '%F %T %Z' '2019-10-11 00:00:00 U' '+%s'
-j is for dry-run; i.e it prevents date from changing system date and time,
-f is for specifying input format,
+%s is for converting given date to seconds since Epoch.
On NetBSD the following will work:
TZ=GMT0 date -d '2019-10-11 00:00:00' '+%s'
Note the use of the TZ environment variable to specify the input timezone instead of trying to have it parsed from the input (though it may be possible to have a more properly formatted timezone parsed from the input, though then that leaves the question of what timezone the output should be formatted in).
On MacOS you might try something similar to what Oguz suggested:
TZ=GMT0 date -j -f '%Y-%m-%d %H:%M:%S' '2019-10-11 00:00:00' '+%s'
I am trying to convert timestamps read from a file from string to date format so that I can find the difference of 2 dates/timestamps. most of the threads/discussions on web show usage of date argument '-d' to convert the string to epoch or to find the difference of two timestamps Find difference between two dates in bash
But it looks like my environment/OS doesn't support -d date argument. Below are the details of my env:
bash --version
GNU bash, version 3.2.52(1)-release (i386-pc-solaris2.10)
Copyright (C) 2007 Free Software Foundation, Inc.
uname -a
SunOS s01***** 5.10 Generic_147148-26 i86pc i386 i86pc
Sample dates read from file:
START_TIME="09/03/16 - 01:04:56"
END_TIME="09/03/16 - 05:10:44"
Code that I have tried
I have tried to mimic the below code from Find difference between two dates in bash
!# /usr/bin/sh
date1="Sat Dec 28 03:22:19 2013"
date2="Sun Dec 29 02:22:19 2013"
date -d #$(( $(date -d "$date2" +%s) - $(date -d "$date1" +%s) )) -u +'%H:%M:%S'
bash test.sh
date: illegal option -- d
usage: date [-u] mmddHHMM[[cc]yy][.SS]
date [-u] [+format]
date -a [-]sss[.fff]
date: illegal option -- d
usage: date [-u] mmddHHMM[[cc]yy][.SS]
date [-u] [+format]
date -a [-]sss[.fff]
test.sh: line 5: - : syntax error: operand expected (error token is " ")
I don't think syntax error on line 5 is the main culprit cause I didnot find option -d in my date's man page.
In response to comments:
>>> date --version
date: illegal option -- version
usage: date [-u] mmddHHMM[[cc]yy][.SS]
date [-u] [+format]
date -a [-]sss[.fff]
>>> date --help
date: illegal option -- help
usage: date [-u] mmddHHMM[[cc]yy][.SS]
date [-u] [+format]
date -a [-]sss[.fff]
>>> echo $0
bash
Even these arguments are not supported. Apologies if I am committing any silly mistake.
Could someone please give me the equivalent of -d for the env details shared above or a way to find the difference between two dates without using -d.
Thanks in advance
awk mktime has a decent chance of existing on your system:
#!/bin/bash
START_TIME="09/03/16 - 01:04:56"
END_TIME="09/03/16 - 05:10:44"
echo -e "$START_TIME\n$END_TIME" |
tr '/:-' ' ' |
awk '{print "20"$3" "$2" "$1" "$4" "$5" "$6}' |
awk '{printf "%s ", mktime($0)}' |
awk '{print $2 - $1}'
explanation:
echo both time strings
tr converts 09/03/16 - 01:04:56 to 09 03 16 01:04:56
first awk changes 09 03 16 01 04 56 to 2016 03 09 01 04 56
second awk converts 2016 03 09 01 04 56 to epoch time: 1457514296 and prints both on one line: 1457514296 1457529044
third awk subtracts first from second, giving difference in seconds: 14748
the awks could also easily be merged, but here i kept each separate for clarity.
According to the POSIX standard, date does not do date and time math for you. It gets or sets the system date and time, possibly with timezone adjustments. See date(1) and note the lack of a -d option (or indeed, any interesting options!).
The question becomes "How do we do date and time math without date?" The timestamps you provided do not have time zone information, so the usual behavior is to assume local time. Local time is bad. Seriously. Some time zones have crazy discontinuities or didn't meaningfully exist hundreds of years ago (e.g. most American time zones). So while you might be able to hack together a custom solution that works in your particular part of the world during the recent past, it simply will not be robust.
If you can get your timestamps into Unix time, you can just subtract them, and that will give you a mostly but not entirely correct answer. Unfortunately, to the best of my knowledge, that can't be done at the command line. Unix provides strptime(3) to do this from C (and from there you go on to mktime(3), as shown in this answer), but I don't believe there is any fully standard utility which provides a command-line interface for this. You may have to write and compile your own.
Correcting for leap seconds is difficult in the extreme because, to the best of my knowledge, POSIX has never provided a standard interface for finding out when leap seconds have happened in the past. Such an interface would require internet connectivity in order to remain up-to-date, which is likely a non-starter for a number of implementations. Without knowing more about your system and its capabilities, I simply cannot begin to guess at what will or will not work for your use case.
I was working with the date command today and discovered some behaviour I cannot explain (I skimmed through: ~$ info '(coreutils) date invocation' but didn't find anything) and hope that someone here might help me to understand why this is happening.
~$ date -u +%F -d "feb 28 -3years"
>> 2012-02-28
~$ date -u +%F -d "feb 29 2012"
>> 2012-02-29
~$ date -u +%F -d "feb 29 -3years"
>> date: invalid date ‘feb 29 -3years’
~$ date -u +%F -d 'feb 29 3 years ago'
>> date: invalid date ‘feb 29 3 years ago’
This is happening on all leap years/leap days (...2008/2012/2016...) and I just can't figure out why.
I' am running above commands on my Ubuntu Gnome Edition (15.04) and on my Server which runs Debian/Jessie.
It's kinda hacky but a good way of getting the last day of the month is to go to the first day of the next month, then back by one day:
$ date -u +%F -d "mar 1 -3years -1day"
2012-02-29
I don't claim to know how the date command is implemented but it stands to reason that feb 29 is the first part of the string that is interpreted, before making the subtractions. So the safest thing to do is to use a date that is always valid, then make a series of subtractions.
I want to get the date +%m/%d/%Y a different number of days ago.
$(( $(date +%s)- 259200)) gives me the seconds of 3 days ago. The -d parameter doesn't accept the seconds to parse it in the format I want 05/06/2011 (error message: date: invalid date '1307284916').
Is there a way to get date to work?
To get date on linux to accept a simple timestamp for -d, prefix the number with the # sign!
~% date +%s
1307548153
~% date -d #1307548153
Wed Jun 8 16:49:13 IST 2011
On MacOS X, you can use date -r 1307284916 +%m/%d/%Y to get the result you want. If you are not on a BSD-derived system like MacOS X (e.g. if you are using Linux), there doesn't seem to be a simple analogue of date -r. And traditional versions of Unix usually don't even provide the facilities of GNU's date.
If you get really stuck, contact me (see my profile) for a program timestamp:
$ timestamp -T '%m/%d/%Y' 1307284916
1307284916 = 06/05/2011
$ timestamp -n -T '%m/%d/%Y' 1307284916
06/05/2011
$ timestamp -T '%m/%d/%Y' 1307284916 $(( $(date +%s)- 259200))
1307284916 = 06/05/2011
1307288704 = 06/05/2011
$