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

Try the following method:

dt=`case "$OSTYPE" in darwin*) date -v-1d "+%s"; ;; *) date -d "1 days ago" "+%s"; esac`
echo $dt

It works on both Linux and OSX.