In a unix shell, how to get yesterday's date into a variable?

Question

I\'ve got a shell script which does the following to store the current day\'s date in a variable \'dt\':

date \"+%a %d/%m/%Y\" | read dt
echo ${dt}
         


        

Answer 1

If you have Perl available (and your date doesn't have nice features like yesterday), you can use:

pax> date
Thu Aug 18 19:29:49 XYZ 2010

pax> dt=$(perl -e 'use POSIX;print strftime "%d/%m/%Y%",localtime time-86400;')

pax> echo $dt
17/08/2010