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
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.
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.
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
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.