Convert string (YYYYMMDD) format to date in Sun OS


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Operating Systems Solaris Convert string (YYYYMMDD) format to date in Sun OS
# 8  
Try a Python approach through bash.
Code:
#!/bin/bash
# epoch.sh
# Usage: ./epoch.sh YYYYMMDD
EPOCH=$1
if [ "$EPOCH" == "" ]
then
	exit 1
fi
echo '#epoch.py
import time
mydate='\"$EPOCH\"'
epoch=int(time.mktime(time.strptime(mydate, "%Y%m%d")))
print(time.strftime("%A", time.localtime(epoch)))
exit()' > "$HOME"/epoch.py
DAY=$( python "$HOME"/epoch.py )
echo "$DAY"

Results OSX 10.13.3, default bash terminal calling Python 2.7.x.
Code:
Last login: Tue Mar 27 22:13:36 on ttys000
AMIGA:amiga~> cd Desktop/Code/Shell
AMIGA:amiga~/Desktop/Code/Shell> ./epoch.sh 20180327
Tuesday
AMIGA:amiga~/Desktop/Code/Shell> ./epoch.sh 20180328
Wednesday
AMIGA:amiga~/Desktop/Code/Shell> _

Note: There is very linited error checking so beware!
EDIT:
This is Python 2.7.10 but I think it works on Python 3.5.2 also.
(I have bent the rules on the Python print statement in 2.7.x AND print() function in 3.5.x...)

Last edited by wisecracker; 03-27-2018 at 07:37 PM.. Reason: See above.
# 9  
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
# 10  
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

# 11  
Quote:
Originally Posted by vgersh99
good idea, but...somehow the zeller function doesn't work correctly:
Code:
$ zeller.sh 2017 3 25
6
$ zeller.sh 2017 3 1
3

The results you are showing seem to indicate that it is working perfectly? What days of the week were you expecting for the 1st and 25th of March last year?
# 12  
Quote:
Originally Posted by Don Cragun
The results you are showing seem to indicate that it is working perfectly? What days of the week were you expecting for the 1st and 25th of March last year?
My bad - was validating against 2018.
Thanks Don!
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