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 #outp         


        

Answer 1

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)

Answer 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

Answer 3

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

Answer 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

Answer 5

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

Answer 6

one liner for mac os x:

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