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 are on a Mac or BSD or something else without the --date option, you can use:

date -r `expr \`date +%s\` - 86400` '+%a %d/%m/%Y'

Update: or perhaps...

date -r $((`date +%s` - 86400)) '+%a %d/%m/%Y'