Extract from cal


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# 8  
Quote:
Originally Posted by protocomm
Excuse me, i don't understand completely the post.
Look at the output from cal 6 2014:
Code:
     June 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

The code you suggested:
Code:
cal 6 2014 | awk 'NR==4{print $1}'

will print 8, but the 1st Sunday in June, 2014 is the 1st; not the 8th.

The code I suggested will print 1 in this case and should print the correct day number for the 1st Sunday no matter what day of the week the 1st day of the month is.
# 9  
Quote:
Originally Posted by Don Cragun
Look at the output from cal 6 2014:
Code:
     June 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

The code you suggested:
Code:
cal 6 2014 | awk 'NR==4{print $1}'

will print 8, but the 1st Sunday in June, 2014 is the 1st; not the 8th.

The code I suggested will print 1 in this case and should print the correct day number for the 1st Sunday no matter what day of the week the 1st day of the month is.
Yes it's true, my answer to the post was so quickly, thx.
# 10  
Are we looking at this the wrong way? The output from cal will have a header row, then the next line will always have Saturday in it. You can see what the Saturday is by trimming off things you don't need, then work out Sunday from that.

Consider:-
Code:
topline=`cal $mm $yyyy | head -3 | tail -1`
Sat1="${topline##* }"
((Sun1=$Sat1+1))
if [ $Sun1 -eq 8 ]
then
   Sun1=1
fi

So, first Saturday of a month is the value of Sat1 and the first Sunday of a month is the value of Sun1

Okay, it's a few lines of code, but it works.


I hope that this helps,
Robin

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

As does:-
Code:
cal $1 $2|awk 'NF == 7 && NR > 2 {print $1; exit}'

....posted by Don earlier that I hadn't fully read. Smilie

Last edited by rbatte1; 08-12-2014 at 12:21 PM.. Reason: Commented on Don's earlier suggestion
# 11  
I think you're onto something! The fourth line of the calendar is always complete, why not calculate from it?

That's robust/complete enough to work on any day of the week. 1 for Sunday, 7 for Saturday.

Code:
$ cal
     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

$ for N in 1 2 3 4 5 6 7
do
        cal | awk 'NR==4 { while(($DAY+0) > 7) $DAY -= 7; print $DAY; exit }' DAY=$N
done

3
4
5
6
7
1
2

$

There's probably a smarter way with modulous, but that's tricky and longwinded when numbers count from 1 instead of 0, so maybe not...
# 12  
Nobody remembers my datecalc script. Smilie

Well, datecalc provides the primatives needed to handle any date calculation. To find the first Sunday...

Code:
#! /bin/ksh
alias datecalc=./datecalc
year=$1
month=$2
day1=$(datecalc -d $year $month 1)
((first_sunday = (8 - day1) % 8 ))
echo $first_sunday
exit 0

Here is a sample run...
Code:
$
$ ./first-sunday 2014 8
3
$

My datecalc script is posted here: datecalc
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