I'm trying desperatly to find a bash or ksh routine that allows me to find for example the previous Monday,Tuesday,Wednesday,... preceding today's date. Additonal it has to work on plain vanilla Solaris X and I don't have the GNU date available.
eg: Today = Thursday 2013/01/17 ; Let's say I want to find the last Monday. It has to return: 2013/01/14
I've managed to find a script on the net that does the job perfectly for all days except in this specific case: eg: Today = Thursday 2013/01/17 ; I want to find the last Thursday which should give as result: 2013/01/10 ; but instead I get todays date again.
The script used was this:
#!/bin/ksh
#Get the nbr of the current weekday (1-7)
DATEWEEK=`date +"%u"`
#Which previous weekday will we need (1-7)
WEEKDAY=$1
# Main part
#Get current date
DAY=`date +"%d"`
MONTH=`date +"%m"`
YEAR=`date +"%Y"`
#Loop trough the dates in the past
COUNTER=0
if [[ $DATEWEEK -eq $WEEKDAY ]] ; then
# I need to do something special for the cases when I want to find the date of the same day last week
DAYS_BACK=168
DAY=`TZ=CST+$DAYS_BACK date +%d`
echo "DAY (eq) = $DAY"
else
while [[ $DATEWEEK -ne $WEEKDAY ]] ; do
COUNTER=`expr $COUNTER + 1`
echo "Counter is: $COUNTER"
DAYS_BACK=`expr $COUNTER \* 24`
echo "DAYS BACK is: $DAYS_BACK"
DAY=`TZ=CST+$DAYS_BACK date +%d`
echo "DAY is: $DAY"
if [[ "$DAY" -eq 0 ]] ; then
MONTH=`expr "$MONTH" - 1`
if [[ "$MONTH" -eq 0 ]] ; then
MONTH=12
YEAR=`expr "$YEAR" - 1`
fi
fi
DATEWEEK=`expr $DATEWEEK - 1`
if [[ $DATEWEEK -eq 0 ]]; then
DATEWEEK=7
fi
done
fi
echo $DAY/$MONTH/$YEAR
