Pattern searching inside Variable - not looking at files

Rediranch · #1 (**permalink**) 06-06-2005

Hi,

I've searched this site and not found this already, so if I missed on my search, sorry.

I need to pass in a variable to a script, where the first three characters of that variable represent a calendar quarter, and the last 2 characters are the year. I.E. Q0105 for Q1, Q0205 for Q2, and so on. How to do this is not the question.

Inside the script, I need to pattern match on the first three characters to determine what quarter it is (technically I could match on the 3rd char only but it's easier to read and understand if I do all three - but not stuck on that).

The reason is I need to populate three other variables based on the quarter. The other variables represent the months, so for Q0105, I need to make M0105, M0205, M0305. For Q0405 I need to make M1005, M1105, M1205.

So, how do I read just the first three chars that are inside the input variable, to match it up to a quarter?

Thanks!

Rediranch · #2 (**permalink**) 06-06-2005

Of course, as soon as I post this, the guy here at work that I asked, who had no idea, came up with something that worked.....

INPUT_VAL=$1

if [[ ${INPUT_VAL%${INPUT_VAL##Q01}} = "Q01" ]]; then

print "Quarter is 1"
MONTH1="M0105"
MONTH2="M0205"
MONTH3="M0305"

else

if [[ ${INPUT_VAL%${INPUT_VAL##Q02}} = "Q02" ]]; then
print "Quarter is 2"
MONTH1="M0405"
MONTH2="M0505"
MONTH3="M0605"

else
if [[ ${INPUT_VAL%${INPUT_VAL##Q03}} = "Q03" ]]; then
print "Quarter is 3"
MONTH1="M0705"
MONTH2="M0805"
MONTH3="M0905"

else
if [[ ${INPUT_VAL%${INPUT_VAL##Q04}} = "Q04" ]]; then
print "Quarter is 4"
MONTH1="M1005"
MONTH2="M1105"
MONTH3="M1205"

else
print "Input was not in quarter format"
fi

fi

fi

fi

vgersh99 · #3 (**permalink**) 06-06-2005

no input validation - taking the set assumptions

red.sh Q0105
red.sh Q0305

red.sh:

Code:

#!/bin/ksh

typeset -Z2 year=$(echo $1 | sed 's/^.*\(..\)$/\1/')
typeset -i quat=$(echo $1 | sed 's/^Q\(..\).*$/\1/')

typeset -Z2 mon

for i in 1 2 3
do
   mon="$(( i + (quat - 1) * 3 ))"
   eval m${i}="M${mon}${year}"
done

echo "m1->[${m1}]"
echo "m2->[${m2}]"
echo "m3->[${m3}]"

which can be simplified a bit further given ksh-nature of the script - left as an exercise.

Rediranch · #4 (**permalink**) 06-07-2005

Thanks, vgersh99.

Not sure what all your statements do, but it does work.

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