show: show all namespaces in "work" view...
show: show all namespaces in "work" view
This commit addresses a number of deficiencies in `hg show work`'s
output:
* Failure to render tags (it just wasn't implemented)
* Failure to render names associated with non-built-in namespaces
(e.g. remotenames)
* Color names were hardcoded instead of coming from the canonical
source in the namespace
This change has the intended effect of rendering tags and extra
namespaces. It solves an immediate need at Mozilla of having
names from a custom namespace printed, which is blocking us from
switching from a custom `hg wip` revset/template combo to `hg show
work`.
Note that the order of branches and bookmarks changes. This is
because bookmarks are registered before branches in namespaces.py.
We may want to register them last, after tags and branches. Or we
may want to added a weighted field to the namespace to control
display order. Something to think about.
I'm not a big fan of the complexity in the templating layer. There
is a lot of code to basically filter out the special case of
branch=='default' and tag=='tip'. Ideally, we would iterate over
a data structure that had irrelevant/unwanted names pre-filtered.
However, I wasn't sure how to best implement this. We probably
want {namespaces} to emit everything (its current behavior). I
was toying with the following:
* {namespacesnondefaults} variation that filtered values
* A filter function that operated on {namespaces} (I wasn't sure
how to implement this since the filtering layer would see a
"hybrid" instance as opposed to something that was definitely
an iterable of namespaces.)
* A namespaces(...) function where you could specify which values
to return. I like this the most. But it really wants named
arguments to control filtering and we only support named arguments
on revsets, not templates.
I figure perfect is the enemy of good and we can refine templating
support for namespaces in the future. At least now we have a
concrete example of a use case.
