I'm new to shell script, and I've for a way to input date without
asking the user to put day, month and year one at a time, any ideas?
$ read -p "Enter date: " d
Enter date: 2015-11-30
$ date -d $d
Mon Nov 30 00:00:00 EST 2015
you can similarly set the date with -s option.
Related
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.
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" "$s"
Thu Sep 21 14:49:51 EDT 2017
You can do something like this:
date -d "Sun Sep 6 02:00:00 IST 2012+10 days"
I'm having trouble with checking time since EPOCH. (and late subtract it from another one).
I get the date like this:
var=$(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
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 am trying to convert a UTC time to GMT time in my small script, but it doesn't work:
TimestampUTC=$(date +"%s")
echo $TimestampUTC
dates=$(date -d #$TimestampUTC)
echo $dates
## 2 hours difference between UTC and GMT
Hours2=120
TimestampGMT=$((TimestampUTC - Hours2))
echo $TimestampGMT
diff=$((TimestampUTC - TimestampGMT))
echo $diff
dateGMT=$(date -d #$TimestampGMT)
echo $dateGMT
The displayed result for $dateGMT is the same as $dates.
Thanks in advance.
error in script.
Unix timestaps are given in seconds.
Hours2=120 means 120 seconds.
So your 2 timestaps are diverging by 2 minutes, not 2 hours.
This code is correct:
Hours2=7200
Also you claim having 2 hours between GMT and UTC, I'm sure you mean CET (central european time)
Note: there is nothing like a CET timestamp. It's just the normal unix timestamp displayed with a timezone offset. So independently of world location, the unix timestamp is always, worldwide, the same at the same instant.
You can replace all your code by just this
# get the timestamp 2 hours in the future from now
date2h=$(date -d "2 hours" +%s)
Which gives you the unix timestamp from the future. It is NOT the current timestamp in CET. The current CET timestamp is always the same as UTC.
How to get the time from UTC and CET? Set the environment variable TZ before the command.
$ TZ=UTC date
Mon Aug 17 11:44:05 UTC 2015
$ TZ=CET date
Mon Aug 17 13:44:05 CEST 2015
$ TZ=GMT date
Mon Aug 17 11:44:05 GMT 2015
but the timestap is always the same
$ TZ=UTC date +%s
1439812072
$ TZ=CET date +%s
1439812072
$ TZ=GMT date +%s
1439812072
GMT and UTC do not differ by 2 hours. In fact they don't differ at all. So displaying the dates of GMT and UTC will always show exactly the same number.
Also I don't know bash but I find it hard to believe that 2 hours is represented by 120 minutes. Normally when doing math with dates milliseconds are used.
In your favourite terminal use the following sequence
export TZ=GMT; date
date_format='+%d %B %Y %H:%M'
datatest="2021-11-21 12:00:00 UTC"
echo $(date -d "$datatest" "$date_format")
datatest="2021-11-21 12:00:00 CET"
echo $(date -d "$datatest" "$date_format")
datatest="2021-11-21 12:00:00 GMT"
echo $(date -d "$datatest" "$date_format")
Out:
21 November 2021 13:00
21 November 2021 12:00
21 November 2021 13:00