view mod_auth_pam/mod_auth_pam.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 57bb2497fadc
children
line wrap: on
line source

-- PAM authentication for Prosody
-- Copyright (C) 2013 Kim Alvefur
--
-- Requires https://github.com/devurandom/lua-pam
-- and LuaPosix

local posix = require "posix";
local pam = require "pam";
local new_sasl = require "util.sasl".new;

function user_exists(username)
	return not not posix.getpasswd(username);
end

function test_password(username, password)
	local h, err = pam.start("xmpp", username, {
		function (t)
			if #t == 1 and t[1][1] == pam.PROMPT_ECHO_OFF then
				return { { password, 0} };
			end
		end
	});
	if h and h:authenticate() and h:endx(pam.SUCCESS) then
		return user_exists(username), true;
	end
	return nil, true;
end

function get_sasl_handler()
	return new_sasl(module.host, {
		plain_test = function(sasl, ...)
			return test_password(...)
		end
	});
end

module:provides"auth";