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

Question

How can I print the date which is a day before current time in Bash?

Answer 1

Try the below code , which takes care of the DST part as well.

if [ $(date +%w) -eq $(date -u +%w) ]; then
  tz=$(( 10#$gmthour - 10#$localhour ))
else
  tz=$(( 24 - 10#$gmthour + 10#$localhour ))
fi
echo $tz
myTime=`TZ=GMT+$tz date +'%Y%m%d'`

Courtsey Ansgar Wiechers

Answer 2

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

Answer 3

date -d "yesterday" '+%Y-%m-%d'

or

date=$(date -d "yesterday" '+%Y-%m-%d')
echo $date

Answer 4

date --date='-1 day'

Answer 5

Advanced Bash-scripting Guide

date +%Y:%m:%d -d "yesterday"

For details about the date format see the man page for date

date --date='-1 day'

Answer 6

You could do a simple calculation, pimped with an regex, if the chosen date format is 'YYYYMM':

echo $(($(date +"%Y%m") - 1)) | sed -e 's/99$/12/'

In January of 2020 it will return 201912 ;-) But, it's only a workaround, when date does not have calculation options and other dateinterpreter options (e.g. using perl) not available ;-)