I know the command date +"%Y%m%d" can change today's date to digit format as below .
$date +"%Y%m%d"
20071217
it works fine .
now I want to do it back . If I have a file like below, (in the file , there are three lines, and each line have ; sign , after the ; sign is the date ) , I want to change the date of the first line to the format Dec 17 , can advise what can i do ? thx
best way is to parse it. if each field after the ; is exactly the same width - ie; first 4 digits are the year, next 2 digits are the month, and last 2 digits are the day, then it will be quite easy for you to convert with awk, or with a cut + date + case.
the last field may pose you issues though: 20071254