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mod_broadcast/mod_broadcast.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 |
parent | 1016:9f7c97e55593 |
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local is_admin = require "core.usermanager".is_admin; local allowed_senders = module:get_option_set("broadcast_senders", {}); local from_address = module:get_option_string("broadcast_from"); local jid_bare = require "util.jid".bare; function send_to_online(message) local c = 0; for hostname, host_session in pairs(hosts) do if host_session.sessions then for username in pairs(host_session.sessions) do c = c + 1; message.attr.to = username.."@"..hostname; module:send(message); end end end return c; end function send_message(event) local stanza = event.stanza; local from = stanza.attr.from; if is_admin(from) or allowed_senders:contains(jid_bare(from)) then if from_address then stanza = st.clone(stanza); stanza.attr.from = from_address; end local c = send_to_online(stanza); module:log("debug", "Broadcast stanza from %s to %d online users", from, c); return true; else module:log("warn", "Broadcasting is not allowed for %s", from); end end module:hook("message/bare", send_message);