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view mod_conformance_restricted/mod_conformance_restricted.lua @ 6029:cc665f343690

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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 7dbde05b48a9
children
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-- Prosody IM
-- Copyright (C) 2012 Florian Zeitz
--
-- This project is MIT/X11 licensed. Please see the
-- COPYING file in the source package for more information.
--

local st = require "util.stanza";
local jid = require "util.jid";

module:hook("message/host", function (event)
	local origin, stanza = event.origin, event.stanza;
	local node, host, resource = jid.split(stanza.attr.to);
	local body = stanza:get_child_text("body");

	if resource ~= "conformance" then
		return; -- Not interop testing
	end

	if body == "PI" then
		origin.send("<?testing this='out'?>");
	elseif body == "comment" then
		origin.send("<!-- no comment -->");
	elseif body == "DTD" then
		origin.send("<!DOCTYPE greeting [\n<!ENTITY test 'You should not see this'>\n]>");
	elseif body == "entity" then
		origin.send("<message type='chat' to='"..stanza.attr.from.."'><body>&test;</body></message>");
	else
		local reply = st.reply(stanza);
		reply:body("Send me one of: PI, comment, DTD, or entity");
		origin.send(reply);
	end

	return true;
end);