Given a UNIX date variable, I need to construct a new date adding a certain number of hours (must be variable) to the initial date. I've looked at this post, which explains how to do this with the current date. For example with 1 hour, date -d '+1 hour' '+%F %T' returns the time it will be in exactly one hour. However, I need to do this with a date variable, not with the current date.
I've tried messing around with the -d flag, but if I set the date to another date variable, I can't figure out how to change it again (such as adding another X number of hours).
Is there a good way to do this? Am I on the right track with the -d flag or is there a better way?
Thanks!
You can use:
# your date variable
dt=$(date -d '2016-08-15 11:10:15')
# add 1 hour to $dt now
date -d "$dt +1 hour"
Mon Aug 15 12:10:15 EDT 2016
Related
I want to store yesterday's date in BASH variable to search for yesterday's files with that variable in the file-name wildcard search. I am using the following format for New York City, NY, USA (EST) time zone, and wanted to know whether it is guaranteed to fetch yesterday's date from the system date; else I can make further changes to the variable.
yesterday=$(TZ=GMT+28 date +%Y%m%d)
...
for file in $HOME_DIR/*$yesterday*.txt;
...
The text filename in HOME_DIR would be as follows:
"ABC_20171011064612.txt"
update 1: Attempt for removing daylight savings related issues:
yesterday=$(echo -e "$(TZ=GMT+28 date +%Y%m%d)\n$(TZ=GMT+18 date +%Y%m%d)"|grep -v $(date +%Y%m%d)|sort|tail -1)
1) Convert two dates to string, 24 hours and 14 (picked arbitrarily to be less than 24 hours) hours before today's date
2) Filter for dates that are not today's date
3) Sort strings from 2) in ascending order
4) Assign yesterday variable to last tail -1 entry of the list
It may not be always right due to DST, although it will not be a big issue.
You could rather say:
yesterday=$(date -d yesterday +%Y%m%d)
You attempted
yesterday=$(echo -e "$(TZ=GMT+28 date +%Y%m%d)\n$(TZ=GMT+18 date +%Y%m%d)|
grep -v $(date +%Y%m%d)|sort|tail -1)
I think it worked.
I'm having trouble with checking time since EPOCH. (and late subtract it from another one).
I get the date like this:
var=$(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
Need a help in unix shell script in calculating date.
I will be getting date value (eg: 20150908) as parameter, now inside the script i need to calculate 7 days ago date (20150908 -7).
something like below:
date=20150908
lastweek_date=20150908 - 7 ---> this should output as 20150901
Could someone help me on this.
Thanks
With GNU date, we can subtract one week:
$ date -d "20150908 - 1 week" '+%Y%m%d'
20150901
Alternatively, we could subtract 7 days:
$ date -d "20150908 - 7 days" '+%Y%m%d'
20150901
And, to show that this works over month boundaries:
$ date -d "20150901 - 1 week" '+%Y%m%d'
20150825
This solution is not OSX/BSD compatible.
A week is 604800 seconds long so to get the number of seconds since the epoch in a portable and POSIX compliant fashion and use it to compute the date 1 week ago do as follows:
PRESENT=$( date +%s )
WEEKAGO=$(( PRESENT - 604800 ))
printf "%s\n" "$WEEKAGO"
I have an input like 2013_07_02 (It may be any date). I want to retrieve date which is 2 days back from the input date. How to do so?
In case you want the output in the same format as the input, i.e. 2013-07-02, substitute _ with - before passing to date:
$ inputdate='2013_07_02';
$ date --date=${mydate//_/-}'-2 day' +'%Y_%m_%d'
2013_06_30
date is your friend:
date -d "2013-01-01 -2 days" +"%Y-%m-%d"
This prints out 2012-12-30.
use date with -d option. It understands relative dates, even with common language like
date -d "two days ago".
In your case something like
date -d "2013-07-01 -2 days"
would be enough.
At work all the days config files are generated fresh and appended with a
session number. The company went public on Feb 16, and the 86400 is seconds
in one day. The session number is generated by subtracting the company start
day from seconds_since_last_day and adding a few zero's
That is the key to interacting with the days config files. I get this - However I do not
understand the
date -ud "$distance days ago 00:00:00".
Is it the number of seconds since 1970?
if $session; then
# return the session of the last day
seconds_since_day_one=`date -ud "Feb 16 2002" +"%s"`
seconds_since_last_day=`date -ud "$distance days ago 00:00:00" +"%s"`
days_between=`printf "%010d" $(( (seconds_since_last_day - seconds_since_day_one) / 86400 ))`
# Truncate on the left to 9 bytes to leave room
# to append the engine suffix for your environment
echo $days_between | awk '{l=length($1); print substr( $1, (l-8), l )}'
date -ud "$distance days ago 00:00:00" in itself just prints the date a certain amount of days ago in a quite readable format, but when you add the FORMAT string to control the output +"%s" does indeed mean the number in so called Unix Time (number of seconds since 1970-01-01 00:00:00 UTC).
If the variable $distance is set to a number it shows the date that number of days ago, if its set to 0 it means today, 1 it means yesterday, 2 the day before yesterday and so on. To better understand these formats and relative keywords there are rather good documentations in (amongst other places) the GNU coreutils package.
Check these URLs:
http://www.gnu.org/software/coreutils/manual/html_node/Relative-items-in-date-strings.html#Relative-items-in-date-strings
http://www.gnu.org/software/coreutils/manual/html_node/Date-input-formats.html
http://www.gnu.org/software/coreutils/manual/html_node/date-invocation.html#date-invocation
Wikipedia explanation of Unix Time:
http://en.wikipedia.org/wiki/Unix_time
The option -d to date provides a generic string to obtain the date.
So, for example, date -d yesterday will print yesterday's date, and date -d 'yesterday 12:00 AM' will print yesterday's date with the time set to 12:00 AM.
So, date -d 6 days ago 00:00:00 will print the date from 6 days ago, with the time set to 00:00:00. I hope it answers your question.
The format +"%s" tells date to print the number of seconds from 1970, instead the date.
mktime and strftime in awk can be used to get the date of the time.
http://www.gnu.org/software/gawk/manual/html_node/Time-Functions.html
For instance, strftime("%A",mktime("YYYY MM DD 00 00 00"))
should give you the day.