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" "$date_format")
Out:
21 November 2021 13:00
21 November 2021 12:00
21 November 2021 13:00
Related
When I run this command I get what you'd expect:
date -d "2018-06-07 + 1 days"
Fri Jun 8 00:00:00 CEST 2018
1 day is added to the day provided (using midnight as starting point).
However when I try to work in a time (17:00:00), two things happen.
date -d "2018-06-07 17:00:00 + 28 days"
Up to 25 days, the output is wrong: wrong dates/wrong time (I have run this in a loop).
Above 25 days, it starts spitting out "date: invalid date ‘2018-06-07 17:00:00 +25 days’"
The manpage says about -d /--date that is pretty much free format. But I'm starting to think the plus sign is incorrectly interpreted (maybe as a timezone offset?) when you use the time (hours:minutes:seconds)?
So how can I add days FROM a timestamped date?
For increment on the days with timestamp to work, the timestamp needs to be in the standard format returned by default by the date command. So sanitize the date to a format in which it accepts minute arithmetic and do the processing.
date -d "2018-06-07 17:00:00"
Thu, Jun 07, 2018 5:00:00 PM
Now put it in a variable, e.g. putting your string in the example below
dateStr=$(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.
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"
It looks like I can't manage to get the bash UTC date in second. I'm in Sydney so + 10hours UTC time
date
Thu Jul 3 17:28:19 WST 2014
date -u
Thu Jul 3 07:28:20 UTC 2014
But when I tried to convert it, I'm getting the same result which is not UTC time
date +%s
1404372514
date -u +%s
1404372515
What am I missing here?
After getting an answer saying date +%s was returning UTC time, here are more details about the problem I'm facing now.
I'm trying to compare a date written in a file with python. This date is written in seconds in UTC time. And the bash date +%s doesn't give me the same one. Actually if I'm doing in python time.asctime(time.localtime(date_in_seconds_from_bash)), I get the current time of Sydney, not UTC. I don't understand.
I believe +%s is seconds since epoch. It's timezone invariant.
I bet this is what was intended as a result.
$ date -u --date=#1404372514
Thu Jul 3 07:28:34 UTC 2014
You say you're using:
time.asctime(time.localtime(date_in_seconds_from_bash))
where date_in_seconds_from_bash is presumably the output of date +%s.
The time.localtime function, as the name implies, gives you local time.
If you want UTC, use time.gmtime() rather than time.localtime().
As JamesNoonan33's answer says, the output of date +%s is timezone invariant, so date +%s is exactly equivalent to date -u +%s. It prints the number of seconds since the "epoch", which is 1970-01-01 00:00:00 UTC. The output you show in your question is entirely consistent with that:
date -u
Thu Jul 3 07:28:20 UTC 2014
date +%s
1404372514 # 14 seconds after "date -u" command
date -u +%s
1404372515 # 15 seconds after "date -u" command
One might consider adding this line to ~/.bash_profile (or similar) in order to can quickly get the current UTC both as current time and as seconds since the epoch.
alias utc='date -u && date -u +%s'
Based on the answer from the other #Adam, here is a one-liner with the UTC date-time and seconds since epoch start:
alias utc='printf "%s : %s\n" "$(date -u)" "$(date -u +%s)"'
The epoch start time is defined as the number of seconds since January 1, 1970 at 00:00 Greenwich Mean Time (GMT), a timezone now known as UTC±00:00 or simply UTC.
How come date is converting to wrong time?
result=$(ls /path/to/file/File.*)
#/path/to/file/File.1361234760790
currentIndexTime=${result##*.}
echo "$currentIndexTime"
#1361234760790
date -d#"$currentIndexTime"
#Tue 24 Oct 45105 10:53:10 PM GMT
This particular timestamp is in milliseconds since the epoch, not the standard seconds since the epoch. Divide by 1000:
$ date -d #1361234760.790
Mon Feb 18 17:46:00 MST 2013
For Mac OS X, it's date -r <timestamp_in_seconds_with_no_fractions>
$ date -r 1553024528
Tue Mar 19 12:42:08 PDT 2019
or
$ date -r `expr 1553024527882 / 1000`
Tue Mar 19 12:42:07 PDT 2019
or
$ date -r $((1553024527882/1000))
Tue Mar 19 12:42:07 PDT 2019
You can use bash arithmetic expansion to perform the division:
date -d #$((value/1000))
Note that "value" is a bash variable with the $ being optional; i.e., $value or value can be used.