convert unix date to readable format

Shell Programming and Scripting
1

Dear Experts,

I need your help to convert a unix date and time format number in to readable format like dd/mm/yyyy .
I have a text file of more than 10,000 records and it is like

NAME DATE1 COUNTRY DATE2
ABD 1223580395699 USA 1223580395699
ABD 1223580395699 USA 1223580395699
ABD 1223580395699 USA 1223580395699
ABD 1223580395699 USA 1223580395699
ABD 1223580394340 USA 1223580387665

I need a unix shell script which will read the records from first line and till end of the file and should convert the unix date to readable format like "ddmmyy".
output should be like this

NAME DATE1 COUNTRY DATE2
ABD 14102008 USA 14102008
NAME 14102008 COUNTRY 14102008

Please tel me how i can achieve this by a unix shell script which can convert any unix datetime like 1223580395699 to a readable format like ddmmyyyy.
Thanks

2

Those are not Unix timestamps; the numbers exceed 32 bits.

1223580395699 is 40743-09-21. (Note: I always use the ISO standard date format.)

This code divides the number by 1,000 to get a valid date, and requires GNU awk:

gawk 'NR == 1 { print; next }
{
  $2 = strftime( "%Y-%m-%d", $2 / 1000 )
  $4 = strftime( "%Y-%m-%d", $4 / 1000 )
  print
}' "$FILE"
3

Hi,

Thnaks for your support but i have tried and it was not working so can you please tell me some other way of doing it.
Is it possible to do this with for loop or any other shell script.
Thanks

4

What does "not working" mean? What did happen?

If the output was incorrect, please post it.

If you received error messages, please post them.