Shell Programming and Scripting

I was trying to get 1st Sunday in a month. I tried using cal [month] [year] followed by awk NF=1 apparently it would give entire 1st field in that month.

Any suggestions

This isn't pretty, but I think it works...

Code:

cal 11 2014 | cut -c1-3 | grep "[0-9]" | head -1
or
foo=`cal 11 2014 | cut -c1-3 | grep "[0-9]" | head -1`
echo ${foo}

Yup it worked. Thanks

try

Code:

cal 12 2014 | awk 'NR==4{print $1}'

That will get the first or second sunday depending on the year. It works for 2014.

So, let's look at what we get from cal (either with no operands, or a month operand and a year operand):

Code:

    August 2014
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

From this we can see that the 1st number on a line with 7 fields (after the header line showing the days of the week) is the 1st Sunday. So, a simple awk shell script for this is:

Code:

#!/bin/ksh
cal $1 $2|awk 'NF == 7 && NR > 2 {print $1; exit}'

If you invoke it with no operands, it will give you the 1st Sunday of the current month:

Code:

./sun1

prints:

Code:

3

If you invoke it with a month and year as operands, it will give you the 1st Sunday of that month in that year:

Code:

./sun1 6 2014

prints:

Code:

1

Many of the other scripts suggested would print 8 for this instead of 1.

As always, if you want to try this on a Solaris/SunOS system, change awk to /usr/xxpg4/bin/awk, /usr/xpg6/bin/awk, or nawk.

Excuse me, i don't understand completely the post.