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mod_admin_probe/mod_admin_probe.lua @ 6057: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> |
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date | Fri, 22 Nov 2024 13:50:48 +0000 (4 months ago) |
parent | 1281:f78661861e98 |
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-- Prosody IM -- Copyright (C) 2014 Florian Zeitz -- -- This project is MIT/X11 licensed. Please see the -- COPYING file in the source package for more information. -- local presence = module:depends("presence"); local send_presence_of_available_resources = presence.send_presence_of_available_resources; local hosts = prosody.hosts; local core_post_stanza = prosody.core_post_stanza; local st = require "util.stanza"; local is_admin = require "core.usermanager".is_admin; local jid_split = require "util.jid".split; module:hook("presence/bare", function(data) local origin, stanza = data.origin, data.stanza; local to, from, type = stanza.attr.to, stanza.attr.from, stanza.attr.type; local node, host = jid_split(to); if type ~= "probe" then return; end if not is_admin(from, module.host) then return; end if 0 == send_presence_of_available_resources(node, host, from, origin) then core_post_stanza(hosts[host], st.presence({from=to, to=from, type="unavailable"}), true); end return true; end, 10);