I know how to format a date in bash using the date command
date -d 20160304 +%Y%m%d
for example. But now I want to pass a date and time and return the hours and minutes. I know the output format I need is +%H%M, but I don't know how to format the date string and it is not in the man pages.
For example if I try any of these:
date -d 201801010500 +%H%M
date -d 20180101_0500 +%H%M
date -d 2018-01-01_0500 +%H%M
date -d 2018:01:01-05:00 +%H%M
I get an "invalid date" error. When I search google I always find answer referring to the output format, not the input format...
GNU date accepts these format, among others I'm sure
$ date -d '2018-02-16 12:34'
Fri Feb 16 12:34:00 EST 2018
$ date -d '2018-02-16T12:34:56'
Fri Feb 16 12:34:56 EST 2018
$ date -d '2018-02-16T12:34:56Z'
Fri Feb 16 07:34:56 EST 2018
In general, can't go wrong with ISO8601 time formats
I'm in Canada/Eastern time zone
The date utility is pretty impressive in making sense of different arguments for the -d option.
Here is just one example:
$ date -d "20180101 05:00:00"
Mon Jan 1 05:00:00 +07 2018
Note +07 is the local timezone.
Related
When I run this command I get what you'd expect:
date -d "2018-06-07 + 1 days"
Fri Jun 8 00:00:00 CEST 2018
1 day is added to the day provided (using midnight as starting point).
However when I try to work in a time (17:00:00), two things happen.
date -d "2018-06-07 17:00:00 + 28 days"
Up to 25 days, the output is wrong: wrong dates/wrong time (I have run this in a loop).
Above 25 days, it starts spitting out "date: invalid date ‘2018-06-07 17:00:00 +25 days’"
The manpage says about -d /--date that is pretty much free format. But I'm starting to think the plus sign is incorrectly interpreted (maybe as a timezone offset?) when you use the time (hours:minutes:seconds)?
So how can I add days FROM a timestamped date?
For increment on the days with timestamp to work, the timestamp needs to be in the standard format returned by default by the date command. So sanitize the date to a format in which it accepts minute arithmetic and do the processing.
date -d "2018-06-07 17:00:00"
Thu, Jun 07, 2018 5:00:00 PM
Now put it in a variable, e.g. putting your string in the example below
dateStr=$(date -d "2018-06-07 17:00:00")
date -d "$dateStr + 28 days"
returns
Thu, Jul 05, 2018 5:00:00 PM
The example uses timezones from IST.
Totally new to BASH. Apologies in advance.
Problem
I'd like to add X days to a specific date.
Code
I figured out that date in BASH retrieves the current date.
I also figured out that I can add X days to the current date in the following way,
expiration_date=$ date -v +1d
which gives,
Tue Sep 26 20:28:13 CEST 2017
which is indeed the date of writing plus X=1 days.
Question
In stead of date in the command line above, I'd like to insert a particular date to which X days will be added, e.g. 20/09/2017.
Don't care about the format of the particular date.
In other words: How do I make the following work,
expiration_date=$ '20/09/2017' -v +1d
Tried this answer, but doesn't do what I want.
Edit: Did not know things are different for OSX.
You can do this way:
dt='2017-09-20'
date -d "$dt +1 day"
Thu Sep 21 00:00:00 EDT 2017
date -d "$dt +2 day"
Fri Sep 22 00:00:00 EDT 2017
It seems OP is using OSX. You can use date addition this way:
s='20/09/2017'
date -j -v +1d -f "%d/%m/%Y" "$s"
Thu Sep 21 14:49:51 EDT 2017
You can do something like this:
date -d "Sun Sep 6 02:00:00 IST 2012+10 days"
time_var="6/23/2016 3:20:00 AM"
(this is in EDT)
We need to get unix timestamp for this variable after converting its value to GMT.
Just use the -u flag while passing the date with -d:
$ time_var="6/23/2016 3:20:00 AM"
$ date -d"$time_var EDT" -u
Thu Jun 23 07:20:00 UTC 2016
Note I also appended EDT to your date.
From man date:
-d, --date=STRING
display time described by STRING, not 'now'
-u, --utc, --universal
print or set Coordinated Universal Time
I'm having trouble with checking time since EPOCH. (and late subtract it from another one).
I get the date like this:
var=$(date)
echo $var
wto, 1 mar 2016, 16:00:14 CET
and later I'm trying to turn it into seconds since epoch:
date -d "$var" +"%s"
date: invalid date ‘wto, 1 mar 2016, 16:00:14 CET’
I'm giving this just as an example. Actually I will be reading the date from file, written in default locale format (I'm operating on couple different machines).
if you type date -h there is the reason why you got this error.
the -d option MUST be declared only with TIME and not with complete DATE format
-d,--date TIME Display TIME, not 'now'
so
date -d "23:59:59"
then:
Tue Mar 1 23:59:59 2016
if you need get only the seconds from a date you have to execute this:
date +"%S"
if you use the -d the output will be deplyed in msec
Am I doing something wrong? I can't believe date -d does not accept its own output when I use the ISO 8601 option with seconds. I either have to remove the 'T' or the timezone to get it to work.
> date -Iseconds
2014-01-14T11:07:57-0800
> date -d "2014-01-14T11:07:57-0800"
date: invalid date `2014-01-14T11:07:57-0800'
> date -d "2014-01-14 11:07:57-0800" # remove the 'T'
Tue Jan 14 11:07:57 PST 2014
> date -d "2014-01-14T11:07:57" # remove the timezone
Mon Jan 13 20:07:57 PST 2014
My current solution is to use date +"%Y-%m-%d %T %z" instead of date -Iseconds but I just thought that was strange.
Perhaps your locale is interfering with the date parsing?
Try specifying the time locale:
$ LC_TIME=C date -d "2014-01-14T11:07:57-0800"
Tue Jan 14 11:07:57 PST 2014
This works for me with GNU date 8.13.
GNU coreutils have only supported ISO 8601 dates as input since version 8.13 (released on 2011-09-08).
Thanks to #John1024 for the reference to
https://unix.stackexchange.com/questions/107750/how-to-parse-iso8601-dates-with-linux-date-command