Convert date formats in bash

On OSX, I'm using -f to specify the input format, -j to not attempt to set any date, and an output format specifier. For example:

$ date -j -f "%m/%d/%y %H:%M:%S %p" "8/22/15 8:15:00 am" +"%m%d%y"
082215

Your example:

$ date -j -f "%d %b %Y" "27 JUN 2011" +%Y%m%d
20110627

#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