Increment a date in a file

You could use perl:

perl -MTime::Piece -pi -e 's/\d{4}-\d\d-\d\d/
 (Time::Piece->strptime($&,"%Y-%m-%d")+24*60*60)->date/ge' file

would increment every date in the file.

I would write:

increment_date() {
    local current=$(grep -oP 'PROCESS DATE =\K.+' file)
    local next=$(date -d "$current + 1 day" +%F)
    sed -i "/PROCESS DATE =/s/$current/$next/" file
}

cat file
increment_date
cat file
increment_date
cat file

PROCESS DATE =2012-02-28
PROCESS DATE =2012-02-29
PROCESS DATE =2012-03-01

This assumes the PROCESS DATE line only occurs once in the file, and it is on a line by itself.

It depends on how sophisticated you need this. If all you need is to change the 01 to a 02, you can use something like

sed -i 's/01/02/' file

or

perl -i -pe 's/01/02/' file

Or, to be on the safe side, do it only if the 01 is at the end of the line:

sed -i 's/01$/02/' file
perl -i -pe 's/01$/02/' file

Both of the above solutions will modify the original file because of the -i flag.

If you need to be able to take actual dates into account and, for example, increment 2014-02-28 to 2014-03-01, you need to either use a full-fledged programming language that can parse dates or play around with the date command:

$ while IFS='=' read -r text date; do 
    echo "$text=$(date -d "$date + 1 day" +%F)"; 
  done < file > new_file

The above will split input lines on =, saving the part before the = as $text and the rest as $date. Then, the $date is fed to the date command which can do fancy date manipulations like add a day and then print it out in the format you want (in this case, the format is %F which means Year-Month-Day). Note that this assumes GNU date.

The echo command prints the contents of $text and then the date incremented by one day. This will fail if you have more than one = on any line of your file.

My very own dateutils provide a command-line tool dadd which goes through a text file and adds something to every date found:

$ dadd -S +1d FILE
PROCESS DATE =2014-08-02