[ACCEPTED]-regex match bash variable-bash

Accepted answer

Score: 10

Using regex matching in bash:

for a in 'Apprentice Historian (Level 1)' 'Historian (Level 4)' 'Master Historian (Level 7)' ; do
    set "$a"
    echo " === $1 ==="
    [[ $1 =~ (Apprentice|Master)?' '?(.*)' ('Level' '[0-9]+')' ]] \
        && echo ${BASH_REMATCH[${#BASH_REMATCH[@]}-1]}
done

The tricky 5 part is to retrieve the correct member from 4 BASH_REMATCH. Bash does not support non-capturing 3 parentheses, therefore Historian is either 2 under 1 or 2. Fortunately, we know it is 1 the last one.

Score: 5

Samples pure shell:

a="Historian (Level 1)"
noParens=${a/ \(*/}
lastWord=${noParens/[A-Za-z]* /}

a="Muster Historian (Level 1)"
noParens=${a/ \(*/}
lastWord=${noParens/[A-Za-z]* /}

(It's the same expressions 1 in both cases, just repeated for easy testing).

Score: 0

Based on "And I don't know how to match 6 on the $1 argument."

Have I understood you 5 correctly if what you are asking for is 4 not whether your regex is correct but rather 3 how to perform the match against the contents 2 of your bash variable?

matched_text=$(echo $yourbashvariablecontainingthetext | sed 's/your_regex/backreference_etc/')

$yourbashvariablecontainingthetext 1 should be your $1

Source: stackoverflow.com

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