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

You have atleast 2 options

1. Use perl:

   perl -e '@T=localtime(time-86400);printf("%02d/%02d/%02d",$T[4]+1,$T[3],$T[5]+1900)'
   

2. Install GNU date (it's in the sh_utils package if I remember correctly)

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

3. Not sure if this works, but you might be able to use a negative timezone. If you use a timezone that's 24 hours before your current timezone than you can simply use date.