Get Day of Week from date

UNIX HP-UX
1

Hi All,

I have date in string format 'YYYY-MM-DD'. I want to know day of the week for this date.
Example. For '2005-08-21' my script should return '0' or Sunday
For '2005-08-22' it should return '1' or Monday

I want piece of code for HP-UX korn shell.

Appreciate reply on this.

2 
#!/bin/ksh

date='2005-08-30'

eval $(echo "${date}" | nawk -F- '{printf("year=%s month=%s day=%s\n", $1, $2, $3)}')

echo "year->[${year}] month->[${month}] day->[${day}]"

cal "${month}" "${year}" | nawk -v day="${day}" '
  FNR > 2 {
    for(i=1; i <= NF; i++)
      if ( $i == day) {
        #printf("day->[%d] row->[%d]\n", $i, FNR-2)
        printf("%d\n", (NF == 7 || FNR!=3) ? i-1 : i+(6-NF))
        exit
      }
  }
'
3

Appreciate vgersh99

4

Can the same logic be rewriten into perl?

5

Change the variable $fmt to get weekday in the format you want I added a comment there:
try -

#!/bin/ksh

# input format YYYY-MM-DD prints day name
dow()  
{
	perl -e '
		use POSIX qw(strftime);
		$fmt = "%A";  # %a = abbreviated weekday %u = weekday number	
		$mday = substr("$ARGV[0]", 8, 2);
		$mon =  substr("$ARGV[0]", 5 ,2);
		$year = substr("$ARGV[0]", 0 ,4);
		$weekday =
		  strftime($fmt, 0, 0, 0, $mday , $mon - 1, $year - 1900, -1, -1, -1);
		print "$weekday";
		' "$1"
}
echo "today is $(date)"
echo "$(dow `date "+%Y-%m-%d"` )"
echo "2007-10-03 was a $(dow "2007-10-03")"
6

Thanks Jim, Its helps!!!:b: