Pattern matching from the input arguments

Question

we're trying to enhance the scripts.

Users will pass some arguments and part of the arguments will have 5.0.3 For an example the input argument would be like Jboss5.0.3GA. Since it ( Jboss5.0.3GA ) has "5.0.3" we'll try to locate the installation binary file "Jboss5.0.3GA.tar".

The current script we've now is a KSH script. I'm trying to use if condition with ksh script

Sample usecases/test:

./test.sh Jboss5.0.3GA
Match found... we'll try to locate the installation binary
./test.sh Jboss5.0.3
Match found... we'll try to locate the installation binary
./test.sh 5.0.3
Match found... we'll try to locate the installation binary
./test.sh Jboss5.1.3
No Match found ... we'll be exiting teh script.

Answer 1

Pattern matching in POSIX shells is done with the case construct. ksh also as the [[ x = pattern ]] operator (also copied by bash and zsh) and [[ x =~ regexp ]] in recent versions.

So:

case $1 in
  (*5.0.3*)
    install=$1.tar
    echo Found;;
  (*)
    echo >&2 Not found
    exit 1;;
esac

Answer 2

I'm not an expert at regular expressions, but this works, at least for what you described.

#!/bin/sh

argument="$1"

#if [[ $argument =~ [a-zA-Z]*5\.0\.3[a-zA-Z]+ ]]; then# only works on bash
if echo $argument | egrep -q '[a-zA-Z]*5\.0\.3[a-zA-Z]+'; then
  #echo "Found: ${BASH_REMATCH[0]}" # for bash
  echo "Match Found"

  # you can check for $argument at some other location, here.

else
  echo "No match"
fi

Saving it as test and running it, gives the following results:

bash test 333xxxx5.0.3xxxxx777
Match Found

bash test 333xxxx5.0.2xxxxx777
No match

bash test 5.0.3xxxxx777
Match Found

bash test 5.0.2xxxxx777
No match

You can add ^ at the beginning and $ at the end, to match the full string or nothing. Like this ^[a-zA-Z]*5\.0\.3[a-zA-Z]+$