Number date to Alphabetic in awk - date

Question

I have a file:
ifile.txt
83080603 55 72
87090607 83 87
88010612 82 44
89080603 55 72
00110607 83 87
01030612 82 44
05120618 84 44
The 1st column shows the date and time with the format YYMMDDHH.
I would like to print the first column as hrHDDMMMYYYY
ofile.txt
03H06Aug1983 55 72
07H06Sep1987 83 87
12H06Jan1988 82 44
and so on
I can't able to convert the digit month-year to text month. My script is
awk '{printf "%s%5s%5s\n",
substr($0,7,2)"H"substr($0,5,2)substr($0,3,2)substr($0,1,2), $2, $3}' ifile.txt

Answer 1

EDIT: Adding solution as per OP's comment for adding year as per condition here.
awk -v curr_year=$(date +%y) '
BEGIN{
num=split("Jan,Feb,Mar,Apr,May,Jun,Jul,Aug,Sep,Oct,Nov,Dec",arr,",")
for(j=1;j<=num;j++){
key=sprintf("%02d",j)
months[key]=arr[j]
}
}
{
year=substr($1,1,2)>=00 && substr($1,1,2)<=curr_year?"20":"19"
$1=substr($1,length($1)-1)"H"substr($1,length($1)-3,2) months[substr($1,3,2)] year substr($1,1,2)
print
year=""
}
' Input_file
Could you please try following, written on mobile and successfully tested it on site https://ideone.com/4zYMu7 this also assume that since your Input_file first 2 letters denote year and you want to print only 19 in case there is another logic to get exact year then please do mention it
awk '
BEGIN{
num=split("Jan,Feb,Mar,Apr,May,Jun,Jul,Aug,Sep,Oct,Nov,Dec",arr,",")
for(j=1;j<=num;j++){
key=sprintf("%02d",j)
months[key]=arr[j]
}
}
{
$1=substr($1,length($1)-1)"H"substr($1,length($1)-3,2) months[substr($1,3,2)] "19" substr($1,1,2)
print
}
' Input_file

Answer 2

With GNU awk:
awk -v OFS=" " '
function totime(t, c, y, m, d, h) {
y = substr(t, 1 ,2)
m = substr(t, 3, 2)
d = substr(t, 5, 2)
h = substr(t, 7, 2)
c = y >= 69 ? 19 : 20 # GNU date treats years 69-99 as 19xx else 20xx
return mktime(c y " " m " " d " " h " 0 0")
}
function transformTime(t) {
return strftime("%HH%d%b%Y", totime(t))
}
{
$1 = transformTime($1)
print
}
' ifile.txt
03H06Aug1983 55 72
07H06Sep1987 83 87
12H06Jan1988 82 44
03H06Aug1989 55 72
07H06Nov2000 83 87
12H06Mar2001 82 44
18H06Dec2005 84 44
Or, Perl
perl -MTime::Piece -lane '
$F[0] = Time::Piece->strptime($F[0], "%y%m%d%H")->strftime("%HH%d%b%Y");
print join(" ", #F);
' ifile.txt

Answer 3

Using any awk in any shell on every UNIX box:
$ cat tst.awk
BEGIN {
split("Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec",mths)
}
{
for (i=1; i<length($1); i+=2) {
t[(i-1)/2+1] = substr($1,i,2)
}
$1 = t[4] "H" t[3] mths[t[2]+0] "19" t[1]
print
}
.
$ awk -f tst.awk file
03H06Aug1983 55 72
07H06Sep1987 83 87
12H06Jan1988 82 44
03H06Aug1989 55 72
07H06Nov1900 83 87
12H06Mar1901 82 44
18H06Dec1905 84 44

Answer 4

{
for (i=0; i<4; i++) {
t[i] = substr($1, 2*i+1, 2);
}
d = t[0]"-"t[1]"-"t[2]" "t[3]":00";
cmd = "date --date='"d"' +%HH%d%b%Y";
cmd | getline $1;
close(cmd);
print
}
Save the above as script.awk and run as awk -f script.awk ifile.txt. It uses the date command for date format conversion.