Subtract days from a date in Bash

Question

I want to subtract "number of days" from a date in Bash. I am trying something like this ..

echo $dataset_date #output is 2013-08-07

echo $date_diff #output is 2   

p_dataset_date=`$dataset_date --date="-$date_diff days" +%Y-%m-%d` # Getting Error

score 68 · Accepted Answer · edited Dec 26 '16 at 03:55

68

You are specifying the date incorrectly. Instead, say:

date --date="${dataset_date} -${date_diff} day" +%Y-%m-%d

If you need to store it in a variable, use $(...):

p_dataset_date=$(date --date="${dataset_date} -${date_diff} day" +%Y-%m-%d)

edited Dec 26 '16 at 03:55

Zombo

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answered Aug 12 '13 at 06:22

devnull

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I had to modify this to `$ date "--date=${dataset_date} -${date_diff} 1 day" +%Y%m%d` to actually subtract the date... otherwise it would have added one day. Is there anything i missed? – Max May 07 '14 at 15:57
@x_mtd Yes, you need to set the variable `date_diff`. Set it to the number of days that you want to subtract. – devnull May 07 '14 at 17:18
Very slight improvement to the command - date --date="${dataset_date} -${date_diff} day" +%Y-%m-%d. Just to make it clearer that the --date is an option to the date command, and the double-quotes are being used to correctly represent the STRING passed to the --date option. – anuragw Aug 24 '16 at 19:37
How to find the difference if dataset_date is in this format: %Y-%m-%d %H:%M:%S – Etisha Apr 30 '20 at 11:24

Jeremy · Answer 2 · 2020-05-18T13:11:01.813

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one liner for mac os x:

yesterday=$(date -d "$date -1 days" +"%Y%m%d")

edited May 18 '20 at 13:11

answered Nov 15 '18 at 19:00

Jeremy

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thank you.. I just needed in seconds format, hence went with thirtydaysold=$(date -d "$date -30 days" '+%s') – Santosh Kumar Arjunan Dec 06 '18 at 09:37

score 9 · Answer 3 · answered Jul 08 '20 at 17:54

9

If you're not on linux, maybe mac or somewhere else, this wont work. you could check with this:

yesterday=$(date  -v-1d    +"%Y-%m-%d")

to get more details, you could also see

man date

answered Jul 08 '20 at 17:54

Ankit Bhardwaj

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getting "date: invalid option -- 'v'" – Reishin May 08 '21 at 14:20
what language is this? – May 31 '21 at 18:21
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it is bash scripting – Ankit Bhardwaj Dec 28 '21 at 12:44

Juto · Answer 4 · 2014-07-15T08:13:09.910

To me, it makes more sense if I put the options outside (easier to group), in case I will want more of them.

date -d "$dataset_date - $date_diff days" +%Y-%m-%d

Where:

 1. -d --------------------------------- options, in this case 
                                         followed need to be date 
                                         in string format (look up on $ man date)
 2. "$dataset_date - $date_diff days" -- date arithmetic, more 
                                         have a look at article by [PETER LEUNG][1]
 3. +%Y-%m-%d -------------------------- your desired format, year-month-day

score 4 · Answer 5 · answered Mar 16 '15 at 11:55

4

Here is my solution:

echo $[$[$(date +%s)-$(date -d "2015-03-03 00:00:00" +%s)]/60/60/24]

It calculates number of days between now and 2015-03-03 00:00:00

answered Mar 16 '15 at 11:55

Paweł Dulęba

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score 4 · Answer 6 · answered May 10 '21 at 20:26

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Here is my solution:

today=$(date +%Y%m%d)
yesterday="$(date -d "$today - 1 days" +%Y%m%d)"
echo $today
echo $yesterday

answered May 10 '21 at 20:26

José Roberto Rivas

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This is in essence answer from @Jeremy – Anton Krug May 11 '21 at 10:48

score 2 · Answer 7 · edited Oct 07 '14 at 10:26

2

Below code gives you date one day lesser

ONE=1
dataset_date=`date`
TODAY=`date -d "$dataset_date - $ONE days" +%d-%b-%G`
echo $TODAY

edited Oct 07 '14 at 10:26

Gergo Erdosi

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answered Oct 07 '14 at 10:10

Subtract days from a date in Bash

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