view mod_unsubscriber/mod_unsubscriber.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 e00dc913d965
children 9c9889ad4302
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assert(module:get_host_type() == "component", "This module should be loaded as a Component");

local st = require "util.stanza";

module:hook("presence/bare", function(event)
	local origin, stanza = event.origin, event.stanza;
	if stanza.attr.type == "probe" then
		-- they are subscribed and want our current presence
		-- tell them we denied their subscription
		local reply = st.reply(stanza)
		reply.attr.type = "unsubcribed";
		origin.send(reply);
		return true;
	elseif stanza.attr.type == nil then
		-- they think we are subscribed and sent their current presence
		-- tell them we unsubscribe
		local reply = st.reply(stanza)
		reply.attr.type = "unsubcribe";
		origin.send(reply);
		return true;
	end
	-- fall trough to default error
end);