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} 

How would i go about getting yesterdays date into a variable?

Basically what i'm trying to achieve is to use grep to pull all of yesterday's lines from a log file, since each line in the log contains the date in "Mon 01/02/2010" format.

Thanks a lot

Answer 1

dt=$(date --date yesterday "+%a %d/%m/%Y") echo $dt 

Answer 2

On Linux, you can use

date -d "-1 days" +"%a %d/%m/%Y"