Problems in Program Day of Week

Shell Programming and Scripting
1

My problem is that i want to calculate the day of the week, i want to change result given as a number for the day.

I mean, the program now do that if the day is Monday is represented by 0, Tuesday by 1.. but i want the word "Monday" as a result. I thought in a bucle if but there is always a mistake and i dont know why.

Thank you

#!/bin/sh

#              ja fe ma ap ma ju ju ag se oc no de
set -A lasts 0 31 28 31 30 31 30 31 31 30 31 30 31


day=$1
month=$2
year=$3

#
# Get Day of Week of Jan 1
dow1=$(cal 1 $year | sed -n '3s/. //gp')
((dow1=7-dow1))

#
#  It is a Leap Year?
leap=0
if ((!(year%100))); then
     ((!(year%400))) && leap=1
else
     ((!(year%4))) && leap=1
fi

#
#  Set number of days of Februray
lasts[2]=28
((leap)) && lasts[2]=29

#
#  calculate day of year
i=0
previous=0
while ((i < mes)) ; do
     ((previous=previous+lasts))
     ((i=i+1))
done
((doy=previous+dia))

#
#  Calculate day of week
((dow = (doy+dow1-1)%7 ))

#echo dow = $dow
echo diadelasemana=$dow


if [$dow -eq 0] then
echo -"Monday"
else
echo "..the same but with Tuesday...etc" 
fi

exit 0
2 
if [$dow -eq 0] then
echo -"Monday"
else
echo "..the same but with Tuesday...etc" 
fi

use CASE

case $dow in
   0) echo -"Monday" ;;
   1) echo -"Tuesday" ;; 
# Add 2 ~6  here.
   *)  COMMAND ;; ## Default action if no PATTERN matched
esac
3

If you have GNU date, you can do this:

#!/bin/sh
date -d $3$2$1 '+%A'

$> ./test 28 10 09
Wednesday
4

Great!!! Thank you so much