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

Use Perl instead perhaps?

perl -e 'print scalar localtime( time - 86400 ) . "\n";'

Or, use nawk and (ab)use /usr/bin/adb:

nawk 'BEGIN{printf "0t%d=Y\n", srand()-86400}' | adb

Came across this too ... insane!

/usr/bin/truss /usr/bin/date 2>&1 | nawk -F= '/^time\(\)/ {gsub(/ /,"",$2);printf "0t%d=Y\n", $2-86400}' | adb