File

mod_email/mod_email.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>
date Fri, 22 Nov 2024 13:50:48 +0000
parent 3836:070faeaf51bc
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module:set_global();

local moduleapi = require "core.moduleapi";

local smtp = require"socket.smtp";

local config = module:get_option("smtp", { origin = "prosody", exec = "sendmail" });

local function send_email(to, headers, content)
	if type(headers) == "string" then -- subject
		headers = {
			Subject = headers;
			From = config.origin;
		};
	end
	headers.To = to;
	if not headers["Content-Type"] then
		headers["Content-Type"] = 'text/plain; charset="utf-8"';
	end
	local message = smtp.message{
		headers = headers;
		body = content;
	};

	if config.exec then
		local pipe = io.popen(config.exec ..
			" '"..to:gsub("'", "'\\''").."'", "w");

		for str in message do
			pipe:write(str);
		end

		return pipe:close();
	end

	return smtp.send({
		user = config.user; password = config.password;
		server = config.server; port = config.port;
		domain = config.domain;

		from = config.origin; rcpt = to;
		source = message;
	});
end

assert(not moduleapi.send_email, "another email module is already loaded");
function moduleapi:send_email(email) --luacheck: ignore 212/self
	return send_email(email.to, email.headers or email.subject, email.body);
end