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

Not very sexy but might do the job:

perl -e 'my ($sec,$min,$hour,$mday,$mon,$year,$wday,$yday,$isdst) = localtime(time - 86400);$year += 1900; $mon+= 1; printf ("YESTERDAY: %04d%02d%02d \n", $year, $mon, $mday)'

Formated from "martin clayton" answer.