Convert date formats in bash

Question

I have a date in this format: "27 JUN 2011" and I want to convert it to 20110627

Is it possible to do in bash?

Answer 1

#since this was yesterday date -dyesterday +%Y%m%d  #more precise, and more recommended date -d'27 JUN 2011' +%Y%m%d  #assuming this is similar to yesterdays `date` question from you  #http://stackoverflow.com/q/6497525/638649 date -d'last-monday' +%Y%m%d  #going on @seth's comment you could do this DATE="27 jun 2011"; date -d"$DATE" +%Y%m%d  #or a method to read it from stdin read -p "  Get date >> " DATE; printf "  AS YYYYMMDD format >> %s"  `date -d"$DATE" +%Y%m%d`      #which then outputs the following: #Get date >> 27 june 2011    #AS YYYYMMDD format >> 20110627  #if you really want to use awk echo "27 june 2011" | awk '{print "date -d\""$1FS$2FS$3"\" +%Y%m%d"}' | bash  #note | bash just redirects awk's output to the shell to be executed #FS is field separator, in this case you can use $0 to print the line #But this is useful if you have more than one date on a line 

More on Dates

note this only works on GNU date

I have read that:

Solaris version of date, which is unable to support -d can be resolve with replacing sunfreeware.com version of date