regex in bash expression

Question

I have 2 questions about regex in bash expression.

1.non-greedy mode

local temp_input=\'\"a1b\", \"d\" , \"45\"\'
if [[ $temp_input =~ \\\".*?\\\         


        

Answer 1

For your first question, an alternative is this:

[[ $temp_input =~ \"[^\"]*\" ]]

For your second question, you can do this:

temp_input=abcba
t=${temp_input//b}
echo "$(( (${#temp_input} - ${#t}) / 1 )) b"

Or for convenience place it on a function:

function count_matches {
    local -i c1=${#1} c2=${#2}
    if [[ c2 -gt 0 && c1 -ge c2 ]]; then
        local t=${1//"$2"}
        echo "$(( (c1 - ${#t}) / c2 )) $2"
    else
        echo "0 $2"
    fi
}

count_matches abcba b

Both produces output:

2 b

Update:

If you want to see the matches you can use a function like this. You can also try other regular expressions not just literals.

function find_matches {
    MATCHES=() 
    local STR=$1 RE="($2)(.*)"
    while [[ -n $STR && $STR =~ $RE ]]; do
        MATCHES+=("${BASH_REMATCH[1]}")
        STR=${BASH_REMATCH[2]}
    done
}

Example:

> find_matches abcba b
> echo "${MATCHES[@]}"
b b

> find_matches abcbaaccbad 'a.'
> echo "${MATCHES[@]}"
ab aa ad