Question

i’m just wondering how can we use awk to do exact matches.

for eg

$ cal 09 09 2009
   September 2009
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



$ cal 09 09 2009 | awk '{day="9"; col=index($0,day); print col }'
17
0
0
11
20
0
8
0

As you can see the above command outputs the index number of all the lines that contain the string/number “9”, is there a way to make awk output index number in only the 4th line of cal output above.??? may be an even more elegant solution?

I’m using awk to get the day name using the cal command. here’s the whole line of code:

     $ dayOfWeek=$(cal $day $month $year | awk '{day='$day'; split("Sunday Monday Tuesday Wednesday Thursday Friday Saturday", array); column=index($o,day); dow=int((column+2)/3); print array[dow]}')

The problem with the above code is that if multiple matches are found then i get multiple results, whereas i want it to output only one result.

Thanks!

Answer 1

Limit the call to index() to only those lines which have your “day” surrounded by spaces:

awk -v day=$day 'BEGIN{split("Sunday Monday Tuesday Wednesday Thursday Friday Saturday", array)} $0 ~ "\\<"day"\\>"{for(i=1;i<=NF;i++)if($i == day){print array[i]}}'

Proof of Concept

$ cal 02 1956
    February 1956
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


$ day=18; cal 02 1956 | awk -v day=$day 'BEGIN{split("Sunday Monday Tuesday Wednesday Thursday Friday Saturday", array)} $0 ~ "\\<"day"\\>"{for(i=1;i<=NF;i++)if($i == day){print array[i]}}'
Saturday

Update

If all you are looking for is to get the day of the week from a certain date, you should really be using the date command like so:

$ day=9;month=9;year=2009; 
$ dayOfWeek=$(date +%A -d "$day/$month/$year")
$ echo $dayOfWeek
Wednesday