shell script help to a newbie

[bashirshaz]

Hi,

im rather new to linux, trying to get my head around scripting.

i have made a shell script. i am trying to make a script that take in a date (DATE_TO_RUN) and find what day of week the date is.

what i need help with is to get the $DAYOFWEEK variable assigned with the value from printf("%d\n", (NF == 7 || FNR!=3) ? i-1 : i+(6-NF))

How to do this? Thanx, all help appriciated.

#!/bin/bash

DATE_TO_RUN='2012-11-05'

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

DAYOFWEEK=-1

cal "${month}" "${year}" | nawk -v day="${day}" '
  FNR > 2 {
    for(i=1; i <= NF; i++)
      if ( $i == day) {

        printf("%d\n", (NF == 7 || FNR!=3) ? i-1 : i+(6-NF))
        exit
      }
  }
'

echo "The date: " $DATE_TO_RUN " is day of week " $DAYOFWEEK

By running this script i get this output:

year->[2012] month->[11] day->[05]
1
The date:  2012-11-05  is day of week  -1

Want to have been:

year->[2012] month->[11] day->[05]
1
The date:  2012-11-05  is day of week  1

[watael]

hi,

couldn't you rely on `date'?

d2run="2012-11-05"
date -d "$d2run" +%u
1