Hi
I want to convert MAY 05 2005 01:15:00PM date format to 2005/05/05 01:15:00PM .
CAn somebody suggest me a code ,I am new to unix shell programming.
Thanks
Arif

one way.......

echo 'MAY 05 2005 01:15:00PM' | awk -f mab.awk

mab.awk:

BEGIN {
  monN=split("JAN FEB MAR APR MAY JUN JUL AUG SEP OCT NOV DEC", months)
  for(i=1; i<=monN; i++) {
    months[months[i]]=i;
    delete months[i];
  }
}

{ printf("%s/%02d/%s %s\n",  $3, months[toupper($1)], $2, $3) }

Thanks
That was perfect .
Arif

date +%Y/%m/%d,%H:%M:%S |sed 's/,/ /g'

HI
Can we convert the date to a 24 hour format .
MAY 05 2005 01:15:00PM date format to 2005/05/05 13:15:00
Thanks Arif

BEGIN {
  monN=split("JAN FEB MAR APR MAY JUN JUL AUG SEP OCT NOV DEC", months)
  for(i=1; i<=monN; i++) {
    months[months[i]]=i;
    delete months[i];
  }
}

function conv2mil(time,   tA, tAnum, pm_am) {
  tAnum=split(time, tA, ":")
  sec=substr(tA[tAnum], 1, length(tA[tAnum])-2)
  pm_am=toupper(substr(tA[tAnum], length(tA[tAnum])-1))
  hour= (pm_am ~ /^P./) ? tA[1] + 12 : tA[1]
  return (hour ":" tA[2] ":" sec)
}

{ printf("%s/%02d/%s %s\n",  $3, months[toupper($1)], $2, conv2mil($4) ) }

Hi vgersh99 - hope you're still here i saw this thread and it really helped me a lot! however i'm having difficulty now figuring out how to convert this date format (May 5 13:01) to yyyymmddhhmm appending the current year (2009).

for sample: (May 5 13:01) will be transformed to 200905051301

the reason i want to have it this way is in order for me to take the time difference of two variables.

I've tried modifying your script by using sprintf, but i failed to make it work

Hope you can help me with this

Many thanks!