How can I increment a number at the end of a string in bash?

Question

Basically i need to create a function where an argument is passed, and i need to update the number so for example the argument would be

version_2 and after the function it would change it to version_3

just increments by one

in java I would just create a new string, and grab the last character update by one and append but not sure how to do it in bash.

updateVersion() {
  version=$1
}

the prefix can be anything for example it can be dog12 or dog_12 and always has one number to update.

after the update it would be dog13 or dog_13 respectively.

Answer 1

updateVersion()
{
    [[ $1 =~ ([^0-9]*)([0-9]+) ]] || { echo 'invalid input'; exit; }     
    echo "${BASH_REMATCH[1]}$(( ${BASH_REMATCH[2]} + 1 ))"
}

# Usage
updateVersion version_11         # output: version_12
updateVersion version11          # output: version12
updateVersion something_else123  # output: something_else124
updateVersion "with spaces 99"   # output: with spaces 100

# Putting it in a variable
v2="$(updateVersion version2)"
echo "$v2"                       # output: version3

Answer 2

Use parameter expansion:

#! /bin/bash
shopt -s extglob
for version in version_1 version_19 version_34.14 ; do
    echo $version
    v=${version##*[^0-9]}
    ((++v))
    echo ${version%%+([0-9])}$v
done

extglob is needed for the +([0-9]) construct which means "one or more digits".