Scribunto should have an interface to get the groups a user is in. Example:
if mw.user.new(frame.args[1]).groups.sysop then return 'Admin' else return 'User' end
This was requested on https://en.wikipedia.org/w/index.php?oldid=639931577#Simpler_process_for_submitting_a_request_for_arbitration
| Subject | Repo | Branch | Lines +/- | |
|---|---|---|---|---|
| Expose user information to Lua | mediawiki/extensions/Scribunto | master | +521 -0 |
| Status | Subtype | Assigned | Task | ||
|---|---|---|---|---|---|
| Open | Feature | None | T66475 Make crosswiki bits and pieces truly global (tracking) | ||
| Open | None | T176787 Automatic infobox implementation | |||
| Open | None | T121470 Central Global Repository for Templates, Lua modules, and Gadgets | |||
| Declined | None | T135845 Convert any module as central or centralisable | |||
| Open | Feature | None | T50176 Requested Scribunto/Lua built-in methods/functions (tracking) | ||
| Open | None | T85419 User groups should be exposed to Scribunto |
Note that any way to get the "current" user (or any data about the "current" user) would be a no-go, as it would break caching horribly. But access to publicly-accessibly data about a user (i.e. what's available from the API list=users) by name doesn't a priori seem like a problem.
Change 188503 had a related patch set uploaded (by Jackmcbarn):
Expose user information to Lua
A related bug ask also the user language code. An english admin could then better help any other little wiki.
https://phabricator.wikimedia.org/T68051
examples : https://fr.wikisource.org/wiki/Module:ControlArgs#Multilingual_examples._Exemples_multilingues
An obvious use case for this is – Each time a flag is granted or revoked, people on every project manually update all relevant Project namespace pages and templates. All this should be updated automatically from a single source of truth.
Just think of the amount of effort that will be saved!