In a unix shell, how to get yesterday's date into a variable?
问题 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 回答1: If you have Perl available (and your date doesn't have nice features like yesterday ), you can use: