calcuate the week number of a given date

ahmedwaseem2000 · #1 (**permalink**) 09-09-2005

Hi All,

can any one help me fix the error in this - i am still a novice in shell programming. I got this code after some googling now the code works with all the dates( as much as i know) except for 08 th and 09th of every month. can any one of you please help me fix this issue?

Thanks in advance.

Quote:

# !/bin/ksh

calcweek() {
if [ $LEAP -eq 0 ] ; then
set -A DIM 0 0 31 59 90 120 151 181 212 243 273 304 334
else
set -A DIM 0 0 31 60 91 121 152 182 213 244 274 305 335
fi
((JDAY=DIM[$MM]+DD))
((WEEK=(JDAY+$1)/7))
echo $YEAR-$WEEK
}

MM=$1
DD=$2
YEAR=$3

MMDD=$MM$DD
typeset -Z2 WEEK

((YEAR%100)) && ((LEAP=!(YEAR%4))) || ((LEAP=!(YEAR%400)))

cal 01 $YEAR |
awk '{getline;getline;print NF;exit}' |
read W1K

case $W1K$LEAP in
70) echo '0101 52 9999 5' ;;
10) echo '0102 XX 9999 4' ;;
20) echo '0103 53 9999 3' ;;
30) echo '0000 00 9999 9' ;;
40) echo '0000 00 1229 8' ;;
50) echo '0000 00 1230 7' ;;
60) echo '0000 00 1231 6' ;;
71) echo '0101 52 1231 5' ;;
11) echo '0102 52 9999 4' ;;
21) echo '0103 53 9999 3' ;;
31) echo '0000 00 9999 9' ;;
41) echo '0000 00 9999 8' ;;
51) echo '0000 00 1229 7' ;;
61) echo '0000 00 1230 6' ;;
esac | read JCUT PWK DCUT ADJ

if [ $MMDD -le $JCUT ] ; then
if [ $PWK = 'XX' ] ; then
((PYEAR=YEAR-1))
((PYEAR%100)) && ((PLEAP=!(PYEAR%4))) || ((PLEAP=!
(PYEAR%400)))
((PWK=52+PLEAP))
fi
echo $((YEAR=YEAR-1))-$PWK
else
if [ $MMDD -ge $DCUT ] ; then
echo $((YEAR=YEAR+1))-01
else
calcweek $ADJ
fi
fi

exit 0

Perderabo · #2 (**permalink**) 09-09-2005

Posix, in their wisdom, decided that a leading zero indicates an octal constant so 08 and 09 are illegal octal constants. This change broke a lot of code. Did you test for month=8 or 9? Also week number = 8 or 9?

Define precisely what you mean by week number. Are the first 7 days of the year always week 1?

Ygor · #3 (**permalink**) 09-09-2005

If you have perl...

Code:

#!/usr/bin/ksh
YEAR=2005
MM=05
DD=08
WEEKNUM=$(perl -e '
   use Time::Local;
   $DATE = timelocal(0, 0, 0, $ARGV[2], $ARGV[1]-1, $ARGV[0]-1900);
   $YEARDAY = (localtime $DATE)[7];
   $WEEKNUM = int($YEARDAY / 7) + 1;
   print $WEEKNUM;
   ' $YEAR $MM $DD)
echo Week number is $WEEKNUM

Test result...

Code:

Week number is 19

Note that you can get the current week number by using...

Code:

date +%W

vino · #4 (**permalink**) 09-09-2005

Ygor, look at this.

Code:

$ date --date="2005/05/08" +%W 
18
$ ./ygor.ksh                   
Week number for 2005/05/08 is 19

and this

Code:

$ date --date="2005/09/09" +%W 
36
$ ./ygor.ksh                   
Week number for 2005/09/09 is 36

Edit:

Code:

$ date --date="2005/12/31" +%W 
52
$ ./ygor.ksh                   
Week number for 2005/12/31 is 53

Why the difference for an old date and same results for today ? Looks like boundary cases are coming into play.

vino

ahmedwaseem2000 · #5 (**permalink**) 09-09-2005

First of all thank you all for the replies!!!!

Perderabo - I have checked for the month 08 and 09 it is working perfectly fo r other dates. it also works for the week 09 and 08 but the issue is only with the day being 09 or 08 any suggestions would be of great help. I have seen your code of sword fish(Gosh) thats a wonderful peice of work. I hope you have a solution for this as well. it is calculated with the ISO way like the first week could start as soon as 28th of dec or as late as 3rd of jan.

Oops!!!! and for perl -i dont have perl

ahmedwaseem2000 · #6 (**permalink**) 09-09-2005

Guys got the solution - i have added this if statement before calculation. please let me know if you find this is not good.

Code:

# !/bin/ksh

calcweek() {
if [ $LEAP -eq 0 ] ; then
   set -A DIM 0 0 31 59 90 120 151 181 212 243 273 304 334
else
   set -A DIM 0 0 31 60 91 121 152 182 213 244 274 305 335
fi
if [ $DD -eq 08 ];then
DD=8
elif [ $DD -eq 09 ];then
DD=9
fi
((JDAY=DIM[$MM]+DD))
echo $DD day $MM mon
echo $JDAY
((WEEK=(JDAY+$1)/7))
echo $YEAR-$WEEK
}

MM=$1
DD=$2
YEAR=$3

MMDD=$MM$DD
typeset -Z2 WEEK

((YEAR%100)) && ((LEAP=!(YEAR%4))) || ((LEAP=!(YEAR%400)))

cal 01 $YEAR |
  awk '{getline;getline;print NF;exit}' |
  read W1K

case $W1K$LEAP in
  70)  echo '0101 52 9999 5' ;;
  10)  echo '0102 XX 9999 4' ;;
  20)  echo '0103 53 9999 3' ;;
  30)  echo '0000 00 9999 9' ;;
  40)  echo '0000 00 1229 8' ;;
  50)  echo '0000 00 1230 7' ;;
  60)  echo '0000 00 1231 6' ;;
  71)  echo '0101 52 1231 5' ;;
  11)  echo '0102 52 9999 4' ;;
  21)  echo '0103 53 9999 3' ;;
  31)  echo '0000 00 9999 9' ;;
  41)  echo '0000 00 9999 8' ;;
  51)  echo '0000 00 1229 7' ;;
  61)  echo '0000 00 1230 6' ;;
esac | read JCUT PWK DCUT ADJ

if [ $MMDD -le $JCUT ] ; then
   if [ $PWK = 'XX' ] ; then
      ((PYEAR=YEAR-1))
      ((PYEAR%100)) && ((PLEAP=!(PYEAR%4))) || ((PLEAP=!
(PYEAR%400)))
      ((PWK=52+PLEAP))
   fi
   echo $((YEAR=YEAR-1))-$PWK
else
   if [ $MMDD -ge $DCUT ] ; then
      echo $((YEAR=YEAR+1))-01
   else
      calcweek $ADJ
   fi
fi

exit 0

Perderabo · #7 (**permalink**) 09-09-2005

Thanks for posting your final solution. I don't recall ever hearing of ISO week numbers before. So I will add this thread to our date arithmetic FAQ article. I found this page which has an on-line calculator for ISO week numbers.

For the record, here is a datecalc based solution. It emulates the behavior of the on-line calculator in that it outputs a year, a week number, and a day of week number...

Code:

#! /usr/bin/ksh
month=$1
day=$2
year=$3
typeset -Z2 fwk

function calcisoweek
{
        integer year month day dow1y mjd1y mjd isoweek
        year=$1
        month=$2
        day=$3
        dow1y=$(datecalc -d $year 1 1)
        ((dow1y=(dow1y+6)%7))
        mjd1y=$(datecalc -j $year 1 1)
        mjd=$(datecalc -j $year $month $day)
        ((isoweek=(mjd - mjd1y - dow1y + 7 + 3 )/7))
        echo $isoweek
        return 0
}

dow=$(datecalc -d $year $month $day)
((!dow)) && dow=7
fwk=$(calcisoweek $year $month $day)
if ((!fwk)) ; then
        ((year=year-1))
        fwk=$(calcisoweek $year 12 31)
fi
echo ${year}-W${fwk}-${dow}
exit 0

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