properly using the $ atom in a parameter substitution

Question

I'm trying to perform a parameter substitution that strips everything from the first - char to the end of a string like so: v0.1-bla-hblah-232 -> v0.1

So I'm using the following script:

#!/usr/bin/env bash
project_version=$(git describe --tags --long)
clean_version=${project_version%-$}
echo "$project_version"
echo "$clean_version"

If I run this, it does not work:

test λ ./revisioncount.sh
v0.1-2-g2ff1e73
v0.1-2-g2ff1e73

Seems that the $ atom is not being picked up by bash. backslashing it does not work either: \$.

Now if I change the substitution line to this:

clean_version=${project_version%%-*}

Then it works:

test λ ./revisioncount.sh
v0.1-2-g2ff1e73
v0.1

How can I use the $ (which matches the end of the string) to achieve the desired effect?

Answer 1

clean_version=${project_version%%-*}

is the correct thing to use as you have found out.

% and %% match at the end by definition, so there is no need for an anchor like $ to say match at the end.

For reference # and ## match at the beginning.

The interesting case is the / expansion, which matches anywhere by default. Here the first character being # or % forces the pattern to match at the start or end.

Answer 2

The bash idiom ${var%%pattern} use a pattern, not a regular expression.

In a pattern, a .* means something different than your intended "anything".
In fact, a simple * does match a run of any character in a pattern.

Additionally, the % type of match is (by its own nature) delimited by $.

So, this:

echo ${project_version%%-*}

Will match the longest match (%% instead of %) of a dash (-) followed by a run of any character until the end of the string inside project_version.

The correct assignment will be, then:

clean_version=${project_version%%-*}