Hello Awk'inas/Sed'ers;

This is keep ringing on my mind for a while, onto play in calender with awk or sed . Given a date, month and Year would like to find out the day corresponding to it. Am still a noob on awk and sed, hence would like to learn it from your responses.

Here it is;

cal 10 1987
       October 1987
Sun Mon Tue Wed Thu Fri Sat
                 1   2   3
4   5   6   7   8   9  10
11  12  13  14  15  16  17
18  19  20  21  22  23  24
25  26  27  28  29  30  31

Try sth like this...

value=1

cal 10 1987 | awk -v var="$value" 'NR==2{n=split($0,P)}
NR>2{for(i=1;i<=7;i++){if(var == $i){print P[7-NF+i]}}}'

Not sure if this is worth the while...:

$ for DAY in {1..31}
   do
      cal 10 1987|awk -v day=$DAY \
          'NR==1{next}
           NR==2 {split ($0, WD); next}
           match ($0, "(^| )"day" ") {print day,WD[ int((4+RSTART-RSTART%2)/3)]}
          '
   done
1 Th
2 Fr

cal 10 1987|awk -v D=$1 'NR==2 {split($0,WD)} R=match($0,"(^| )"D" ") {print WD[int((4+R-R%2)/3)]}'

You might find the thread Find Day of Week useful.

Pamu, great work... Can you please explain me about the P in split command and the piece P[7-NF+i]

NR==2{n=split($0,P)} means when input record is 2nd (Days of Week) divide all of them and store in an array variable named P

    October 1987
Su Mo Tu We Th Fr Sa
             1  2  3
 4  5  6  7  8  9 10
11 12 13 14 15 16 17
18 19 20 21 22 23 24
25 26 27 28 29 30 31

Sorry bipinajith, one more query. I understood they are splitting the columns and storing each value into the array P. Whats the need of n here n=split($0,P).

---------- Post updated at 07:42 AM ---------- Previous update was at 07:31 AM ----------

@RudiC: Thanks for your code, can you please explain me on "(^| )"D" "
How this works.

@Don: I am looking after your thread