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Awk convert from day of year to date

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# 1  
Old 07-17-2011
Awk convert from day of year to date
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.
Code:
#!/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:
Code:
175    197 17.07.2011 17.Jul.2011 0720

Any help is greatly appreciated
Last edited by Franklin52; 07-17-2011 at 06:31 AM.. Reason: Please use code tags for data and code samples, thank you
# 2  
Old 07-17-2011
What in your result do you want output differently?
# 3  
Old 07-17-2011
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  
Old 07-17-2011
Can you give an example from your data file?
# 5  
Old 07-17-2011
Sure, please find attached.
file1.txt (53.1 KB) 
# 6  
Old 07-17-2011
What does lines like these refer to?

S5013 2194 175
S5009 2194 175
# 7  
Old 07-17-2011
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:
Code:
mindate=strftime("%d.%m.%Y",systime()-(todayJD-minJD)*(60*60*24))
maxdate=strftime("%d.%m.%Y",systime()-(todayJD-maxJD)*(60*60*24))

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