How do I add X days to date and get new date?

I have Linux ( RH 5.3) machine

I need to add/calculate 10 days plus date so then I will get new date (expiration date))

for example

 # date 
 Sun Sep 11 07:59:16 IST 2012

So I need to get

     NEW_expration_DATE = Sun Sep 21 07:59:16 IST 2012

Please advice how to calculate the new expiration date ( with bash , ksh , or manipulate date command ?)

Asked By: yael

||

You can use “+x days” as format string:

$ date -d "+10 days"
Answered By: quanta

You can just use the -d switch and provide a date to be calculated

date
Sun Sep 23 08:19:56 BST 2012
NEW_expration_DATE=$(date -d "+10 days")
echo $NEW_expration_DATE
Wed Oct 3 08:12:33 BST 2012 
  -d, --date=STRING
          display time described by STRING, not ‘now’

This is quite a powerful tool as you can do things like

date -d "Sun Sep 11 07:59:16 IST 2012+10 days"
Fri Sep 21 03:29:16 BST 2012

or

TZ=IST date -d "Sun Sep 11 07:59:16 IST 2012+10 days"
Fri Sep 21 07:59:16 IST 2012

or

prog_end_date=`date '+%C%y%m%d' -d "$end_date+10 days"`

So if $end_date = 20131001 then $prog_end_date = 20131011.

Answered By: user591

In Mac OS (and maybe other BSD distros):

$ date -v -1d

In order to get 1 day back date using date command:

$ date -v -1d

It will give (current date -1) means 1 day before .

$ date -v +1d

This will give (current date +1) means 1 day after.

Similarly below written code can be used in place of "d" to find out year,month etc.

y-Year
m-Month 
w-Week 
d-Day 
H-Hour 
M-Minute  
S-Second
Answered By: Abhilash Kumar

As previous answers, you can use the -d switch like :

date -d "$myDate + 10 days"

But you need to be careful with the time zone!

An example of something that can go wrong :
If you want to get an interval for a day like [start of day, end of day] for the "8 march 2020" with the timezone at Montreal, where the DST start at 2:00:00 am, you could do

startDate=$(date --iso-8601=n -d "8 march 2020 00:00:00.000000001")
endDate=$(date --iso-8601=n -d "$startDate + 1 day")

but instead of

startDate → "2020-03-08T00:00:00,000000001-0500"
endDate   → "2020-03-09T00:00:00,000000001-0400"

you’ll get 1 hour of the 9 march 2020

startDate → "2020-03-08T00:00:00,000000001-0500"
endDate   → "2020-03-09T01:00:00,000000001-0400"

To fix that, you could do the following :

# Returns the difference between two dates' local time offsets. E.g. : "3600" or "- 1800"
getLocalTimeOffsetDiffInSeconds() {

    # Get the offset in the following format: -0530 for -05h30min
    localOffset1=$(date -d "$1" +%z)
    localOffset2=$(date -d "$2" +%z)
    
    # Get the offset in seconds
    localOffsetInSec1=$(echo $localOffset1 | sed -E 's/^([+-])(..)(..)/scale=2;01(2 * 3600 + 3 * 60)/' | bc)
    localOffsetInSec2=$(echo $localOffset2 | sed -E 's/^([+-])(..)(..)/scale=2;01(2 * 3600 + 3 * 60)/' | bc)
    
    # Compute diff
    diffOffsetInSec=$(echo "$localOffsetInSec2 - $localOffsetInSec1" | bc)
    
    # Add a space between the sign and the value, so it can be used by the command "date".
    echo "$diffOffsetInSec" | sed -E 's/([+-])(.*)/1 2/'
}

  startDate=$(date --iso-8601=n -d "8 march 2020 00:00:00.000000001")
  endDate=$(date --iso-8601=n -d "$startDate + 1 day")
  
  # Fixes endDate if the DST changed between the two dates.
  
  dstDiff=$(getLocalTimeOffsetDiffInSeconds "$startDate" "$endDate")
  
  # dateFix will be a string to remove the dstDiff from the date in a format
  # that can be understood by the command "date". E.g. "3600 seconds" or "- 3600 seconds".
  dateFix=$(echo "$(echo "- $dstDiff" | bc | sed -E 's/([+-])(.*)/1 2/')")
  endDate=$(date --iso-8601=n -d "$endDate $dateFix seconds")
Answered By: Adrian B.

I gave up on the GNU date command due to timezone confusion, and went to python.

python -c "import datetime, dateutil.parser; print((dateutil.parser.parse("$myDate") + datetime.timedelta(days=10)).isoformat())"

Maybe it’s overkill, but it’s straightforward to read, understand and modify as needed.

dateutil.parser.parse is not part of the python standard library, so you have to pip install dateutil; alternatively you can use datetime.datetime.fromisoformat, but you need to be more careful about formats.

Answered By: Paul Price
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