Bash - Increment the version number by 1 [closed] - bash

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I'm new to bash script. I need to bump up the version number of my build by 1.
For eg:- if my version is 1.0.0+1, bumping it would generate 1.0.1+2, ie basically incrementing the number by 1 before and after +.
How can this be done in bash script?

Answer 1

The bash approach which is basically close to Python style is to use a regular expression to separate the version number and the rest of the version string:
version=1.0.0+1
if [[ $version =~ ^([^+?]*[.])([0-9]*)[+]([0-9]+) ]]
then
# ... see below
else
echo Unexpected version string: "$version"
fi
When you have a match, you can refer to the capture groups in the regexp. In Python you would use something like result.group(1) to access the groups. In bash, you have for this purpose a single, predefined array BASH_REMATCH, which always refers to the most recent regexp matching. In our case, we would have
version_string_head=${BASH_REMATCH[1]} # 1.0.
version_count_hi=${BASH_REMATCH[2]} # 0
version_count_lo=${BASH_REMATCH[3]} # 1
The rest is trivial:
new_version_string=${version_string_head}$((version_count_hi + 1))+$((version_count_lo + 1))