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;
    };
});