Convert string (YYYYMMDD) format to date in Sun OS Post: 303015166

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Operating Systems Solaris Convert string (YYYYMMDD) format to date in Sun OS Post 303015166 by vgersh99 on Thursday 29th of March 2018 11:13:59 AM

03-29-2018

Moderator

Quote:

Originally Posted by apmcd47

This bash-only script works with bash 3 on Solaris and bash 4 on Ubuntu.

To use it call it with yyyymmdd date strings as arguments.

Code:

#!/bin/bash

zeller() {
   year=$1 month=$2 day=$3
   (( 10#$month < 3 )) && year=$(( year - 1 ))
   F=$(( (10#$day + ((13 * ((10#$month+9) % 12) + 12)/5) + \
                (5*(year%100))/4 + (year/400) - 2*(year/100))%7 ))
   (( F<0 )) && (( F = F + 7 ))
   echo $F
}

for date in $@
do
   y=${date:0:4} m=${date:4:2} d=${date:6:2}
   dow=$(zeller $y $m $d)
   if (( dow == 3 ))
   then 
      printf "%02d/%02d/%4d (%s) is a Wednesday\n" $d $m $y $date
      # US date format: (comment above and uncomment below if in the USA)
      #printf "%20d/%02d/%4d (%s) is a Wednesday\n" $m $d $y $date
   else
      printf "%02d/%02d/%4d (%s) is NOT a Wednesday\n" $d $m $y $date
      # US date format:
      #printf "%02d/%02d/%4d (%s) is NOT a Wednesday\n" $m $d $y $date
   fi
done

The function zeller uses the Zeller algorithm to find the day of week from the given year/month/day. Result: 0 for a Sunday through to 6 for a Saturday.

Andrew

good idea, but...somehow the zeller function doesn't work correctly:

Code:

$ zeller.sh 2017 3 25
6
$ zeller.sh 2017 3 1
3

vgersh99

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