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

Though all good answers, unfortunately none of them worked for me. So I had to write something old school. ( I was on a bare minimal Linux OS )

$ date -d @$( echo $(( $(date +%s)-$((60*60*24)) )) )

You can combine this with date's usual formatting. Eg.

$ date -d @$( echo $(( $(date +%s)-$((60*60*24)) )) ) +%Y-%m-%d

Explanation : Take date input in terms of epoc seconds ( the -d option ), from which you would have subtracted one day equivalent seconds. This will give the date precisely one day back.