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Get the date (a day before current time) in Bash

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How do I subtract one day from a date in Unix?

The easiest way is to convert the date to a unix time_t value (i.e. seconds since the beginning of the epoch, or '1-1-1970 00:00:00'), and then substract 30 days * 86400 seconds per day from that number. e.g. the following example uses set -x so that you can see the value of the D variable as it changes.


if you have GNU date and i understood you correctly

$ date +%Y:%m:%d -d "yesterday"
2009:11:09

or

$ date +%Y:%m:%d -d "1 day ago"
2009:11:09

If you have BSD (OSX) date you can do it like this:

date -j -v-1d
Wed Dec 14 15:34:14 CET 2011

Or if you want to do date calculations on an arbitrary date:

date -j -v-1d -f "%Y-%m-%d" "2011-09-01" "+%Y-%m-%d"
2011-08-31

date --date='-1 day'

MAC OSX

For yesterday's date:

date -v-1d +%F

where 1d defines current day minus 1 day. Similarly,

date -v-1w +%F - for previous week date

date -v-1m +%F - for previous month date

IF YOU HAVE GNU DATE,

date --date="1 day ago"

More info: https://www.cyberciti.biz/tips/linux-unix-get-yesterdays-tomorrows-date.html