Mercurial > prosody-modules
view mod_http_user_count/mod_http_user_count.lua @ 6029:cc665f343690
mod_firewall: SUBSCRIBED: Flip subscription check to match documentation
The documentation claims that this condition checks whether the recipient is
subscribed to the sender.
However, it was using the wrong method, and actually checking whether the
sender was subscribed to the recipient.
A quick poll of folk suggested that the documentation's approach is the right
one, so this should fix the code to match the documentation.
This should also fix the bundled anti-spam rules from blocking presence from
JIDs that you subscribe do (but don't have a mutual subscription with).
| author | Matthew Wild <mwild1@gmail.com> |
|---|---|
| date | Fri, 22 Nov 2024 13:50:48 +0000 |
| parents | a45f2f79e99b |
| children |
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local it = require "util.iterators"; local jid_split = require "util.jid".prepped_split; module:depends("http"); local function check_muc(jid) local room_name, host = jid_split(jid); if not hosts[host] then return nil, "No such host: "..host; elseif not hosts[host].modules.muc then return nil, "Host '"..host.."' is not a MUC service"; end return room_name, host; end module:provides("http", { route = { ["GET /sessions"] = function () return tostring(it.count(it.keys(prosody.full_sessions))); end; ["GET /users"] = function () return tostring(it.count(it.keys(prosody.bare_sessions))); end; ["GET /host"] = function () return tostring(it.count(it.keys(prosody.hosts[module.host].sessions))); end; ["GET /room/*"] = function (request, room_jid) local name, host = check_muc(room_jid); if not name then return "0"; end local room = prosody.hosts[host].modules.muc.rooms[name.."@"..host]; if not room then return "0"; end return tostring(it.count(it.keys(room._occupants))); end; }; });
