How to format date with in the file

Question

Hi,
I have the below data in a file,
101 02100002111406893401207310900A094101xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
5200xxxxxxxxxx D18000_1 CCDXXXXXXX JUL 31201207 1140689340000001
622113010547999999999003 000333333334RE 00030137 onee SYSTEM 0140689340000001
62200000 0000999999999001 0000011111CR460 00030135 twoo system 0140689340000002
82000000040015507087000000000000000003571110D18000_1 140689340000001
Can some one tell me how to read the date which is there in line 1 which is starting with indicator 1 at position 24 to 29
in this example it is 120731
which is in YYMMDD format
I want to replace from 70 to 75 with above date + 2 as 120802 in 2nd line which starts with indicator 5
My output should look like below in a file again
101 02100002111406893401207310900A094101xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
5200xxxxxxxxxx D18000_1 CCDXXXXXXX JUL 31120802 1140689340000001
622113010547999999999003 000333333334RE 00030137 onee SYSTEM 0140689340000001
62200000 0000999999999001 0000011111CR460 00030135 twoo system 0140689340000002
82000000040015507087000000000000000003571110D18000_1 140689340000001
Can some one please tell me how to do this?
Appreciate your response.

Answer 1

The following will help you to extract the date and perform an arithmetic function on a date:
olddate=$(head -1 myfile | cut -c 24-29)
newdate=$(date -d "20${olddate} + 2 days" "+%y%m%d")You may have to implement some logic if the year is before this century, e.g 1999. For the rest I suggest you look into awk or sed to perform a string replacement based on character position.
Edited by: Dude on Aug 7, 2012 12:57 PM