Karthik Nayak <[email protected]> writes:
> Currently the 'lstrip=<N>' option only takes a positive value '<N>'
> and strips '<N>' slash-separated path components from the left. Modify
> the 'lstrip' option to also take a negative number '<N>' which would
> only _leave_ behind 'N' slash-separated path components from the left.
"would only leave behind N components from the left" sounds as if
the result is A/B, when you are given A/B/C/D/E and asked to
lstrip:-2. Given these two tests added by the patch ...
> +test_atom head refname:lstrip=-1 master
> +test_atom head refname:lstrip=-2 heads/master
... I somehow think that is not what you wanted to say. Instead,
you strip from the left as many as necessary and leave -N
components that appear at the right-most end, no?
> --- a/Documentation/git-for-each-ref.txt
> +++ b/Documentation/git-for-each-ref.txt
> @@ -98,7 +98,8 @@ refname::
> abbreviation mode. If `lstrip=<N>` is appended, strips `<N>`
> slash-separated path components from the front of the refname
> (e.g., `%(refname:lstrip=2)` turns `refs/tags/foo` into `foo`.
> - `<N>` must be a positive integer.
> + if `<N>` is a negative number, then only `<N>` path components
> + are left behind.
I think positive <N> is so obvious not to require an example but it
is good that you have one. The negative <N> case needs illustration
more than the positive case. Perhaps something like:
(e.g. %(refname:lstrip=-1) strips components of refs/tags/frotz
from the left to leave only one component, i.e. 'frotz').
Would %(refname:lstrip=-4) attempt to strip components of
refs/tags/frotz from the left to leave only four components, and
because the original does not have that many components, it ends
with refs/tags/frotz?
I am debating myself if we need something like "When the ref does
not have enough components, the result becomes an empty string if
stripping with positive <N>, or it becomes the full refname if
stripping with negative <N>. Neither is an error." is necessary
here. Or is it too obvious?
