CODEWITHSUNDEEP
bashshell
user2488578
user2488578

Reputation: 916

Shell script : Finding year is wrongly printing value

I am trying to find the previous month's year value. Below is my code. Instead of printing 2015, it is wrongly printing as 223. What could be the issue?

findPrvYear(){
echo "Finding Previous Year"
(
set `date +%m" "%Y`
CURMTHY=$1
CURYRY=$2

if [ $CURMTHY -eq 1 ]
then PRVMTHY=12
     PRVYRY=`expr $CURYRY - 1`
else PRVMTHY=`expr $CURMTHY - 1`
     PRVYRY=$CURYRY
fi
echo $PRVYRY  //This is properly printing as 2015
return "$PRVYRY"
)
}

thisMonthY=$(date +%m)
thisYearY=$(date +%y)
findPrvYear $thisMonthY $thisYearY
retPrvYear=$?
echo $retPrvYear  //This is wrongly printing as 223

Upvotes: 0

Views: 37

Answers (1)

sideshowbarker
sideshowbarker

Reputation: 88296

You need to use $(...) to get the output from that function. Not $?, which gives you the return code (which is something different). So, do something more like this:

findPrvYear(){
echo "Finding Previous Year"
(
set `date +%m" "%Y`
CURMTHY=$1
CURYRY=$2

if [ $CURMTHY -eq 1 ]
then PRVMTHY=12
     PRVYRY=`expr $CURYRY - 1`
else PRVMTHY=`expr $CURMTHY - 1`
     PRVYRY=$CURYRY
fi
return "$PRVYRY"
)
}

thisMonthY=$(date +%m)
thisYearY=$(date +%y)
retPrvYear=$(findPrvYear $thisMonthY $thisYearY)

Upvotes: 1

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