Awk convert from day of year to date

Shell Programming and Scripting
1

Hello everyone,
I have a file which has day of year in one of the columns (JD=substr($0,72,3)). The bellow scripts shows me the minimum and maximum values of the JD and I would like to convert the JD to date.

#!/bin/gawk -f
  {
 check=substr($0,1,1)
 if (check == "S") 
 {
  JD=substr($0,72,3);
 {
  if(minJD=="")
 {
  minJD=maxJD=JD
 }
  if(JD>maxJD) 
 {
  maxJD=JD
 }
  if(JD< minJD) 
 {
  minJD=JD
 }
  } 
}
}
END {
today = strftime("%d.%m.%Y")
today2 = strftime("%d.%h.%Y")
time=strftime("%H%M")
print minJD, "  ",maxJD,today,today2,time
 }

Result:

175    197 17.07.2011 17.Jul.2011 0720

Any help is greatly appreciated

2

What in your result do you want output differently?

3

I would like to get the date (day.month.year) calculated for JD=175 (24.Jun.2011) and for JD=197 (16.Jul.2011).

4

Can you give an example from your data file?

5

Sure, please find attached.

6

What does lines like these refer to?

S5013 2194 175
S5009 2194 175

7

S is a code, the next 4 digits represents line, the next 4 are the points (the values of a grid) and the last 3 is the day of the year-JD.

---------- Post updated at 03:53 PM ---------- Previous update was at 03:18 AM ----------

Thanks for your reply, I managed to solve the issue by adding the following code:

mindate=strftime("%d.%m.%Y",systime()-(todayJD-minJD)*(60*60*24))
maxdate=strftime("%d.%m.%Y",systime()-(todayJD-maxJD)*(60*60*24))