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CASE statement

 
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# 1  
Old 12-15-2008
CASE statement
Hi,

I am writing a bash shell script. My script has a few user defined parameters. When the script runs the first thing it does is make sure that these parameters are valid. One of the parameters is called YEAR. A valid input for YEAR can be 1997-2000.

One way I have come up with to ensure that YEAR is valid is by using a case statement:

Code:
case $YEAR in 
    1997 | 1998 | 1999 | 2000 )
    echo  '- YEAR is valid'
    ;;
    *) 
    echo; echo; echo 'ERROR: YEAR is not valid (Must be 1997,1998,1999, or 2000)'
    exit 1
    ;;
esac

However, that is a simple case! In reality YEAR is valid for 1997-2008. Instead of listing each valid year in the case statement is there a way that I could just place all the valid years into an array and use that? ie:

Code:
YEAR_ARRAY=(1997 1998 1999 2000)

case $YEAR in 
    ${YEAR_ARRAY[@]} )
    echo  '- YEAR is valid'
    ;;
    *) 
    echo; echo; echo 'ERROR: YEAR is not valid (Must be 1997,1998,1999, or 2000)'
    exit 1
    ;;
esac

I have tried that and it doesn't work. Am I close? Is there a better way? Thanks a lot.

Mike
# 2  
Old 12-15-2008
Hi.

Here are some related methods, some use seq, all use eval:
Code:
#!/bin/bash -

# @(#) s2       Demonstrate feature.

echo
echo "(Versions displayed with local utility \"version\")"
version >/dev/null 2>&1 && version "=o" $(_eat $0 $1) seq
set -o nounset

echo
echo " Results:"
years="1997|1998|1999|2000|2001"

echo " With a variable:"
for i in 1990 1997 200x 2001
do
  echo " Looking for $i"
  eval "
  case $i in
    $years) echo found $i ;;
    *) echo unknown: $i ;;
  esac
  "
done

echo
echo " With command substitution:"
for i in 1990 1997 200x 2001
do
  echo " Looking for $i"
  eval "
  case $i in
    $(seq -s"|" 1997 2003)) echo found $i ;;
    *) echo unknown: $i ;;
  esac
  "
done

echo
years="$(seq -s"|" 1997 2003)|2007|2008"
echo " With combination:"
for i in 1990 1997 200x 2001 2007
do
  echo " Looking for $i"
  eval "
  case $i in
    $years) echo found $i ;;
    *) echo unknown: $i ;;
  esac
  "
done

exit 0

Producing:
Code:
% ./s2

(Versions displayed with local utility "version")
Linux 2.6.11-x1
GNU bash 2.05b.0
seq (coreutils) 5.2.1

 Results:
 With a variable:
 Looking for 1990
unknown: 1990
 Looking for 1997
found 1997
 Looking for 200x
unknown: 200x
 Looking for 2001
found 2001

 With command substitution:
 Looking for 1990
unknown: 1990
 Looking for 1997
found 1997
 Looking for 200x
unknown: 200x
 Looking for 2001
found 2001

 With combination:
 Looking for 1990
unknown: 1990
 Looking for 1997
found 1997
 Looking for 200x
unknown: 200x
 Looking for 2001
found 2001
 Looking for 2007
found 2007

Best wishes ... cheers, drl
# 3  
Old 12-15-2008
Hammer & Screwdriver What about creating a file with valid years and then grep?
Perhaps something like this might be useful?

create a file of valid years
Code:
> cat file115
1999
2000
2001
2002
2003
2004
2005
2006
2007
2008

input a valid year, and see results of grep
Code:
> echo "2005" | egrep -f file115
2005

input an invalid year, and nothing is returned
Code:
> echo "2009" | egrep -f file115

# 4  
Old 12-15-2008
How about something like this using an if statement?

Code:
#! /bin/bash
YEAR_INPUT=$1
YEAR_START=1997            # Begin year
YEAR_CURR=$(date +%Y)   # End year (current year)

if [ $YEAR_INPUT -ge $YEAR_START -a $YEAR_INPUT -le $YEAR_CURR ]
  then
    echo "valid year"
  else
    echo "invalid year"
fi

 
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