HexChat - Simplest way to send a msg on other user join, either in
channel or msg.
reddit / irc

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mjh215 · 2019-06-14 11:07

I haven't bothered with scripts/scripting for years and I maintain a
few legacy channels that on rare occasion get users popping back in
after years. I get sick of missing them and not able to send them
contact information or get theirs.

Some of the channels I'd just like to have a public message saying to
(people ignore topics) msg me, others I'd like to set specific nicks
for me to msg them.

Thanks.

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levelfield · 2019-06-15 12:14

Does the network you're on have a MemoServ? That would make things
easier. But it's a pretty simple script, so here you go (save as
.lua):


    hexchat.register("Auto-message", "1.0", "Auto-messages certain
nicks/channels")

    local msg = "According to all known laws of aviation, there is no
way that a bee should be able to fly."

    local nicks = {
        freenode = {
                ['##test4910'] = true,
                ['##test4911'] = {
                        testnick1 = true,
                        testnick2 = true
                }
        }
    }


    local function spam(nick)
        hexchat.command("MSG " .. nick .. " " .. msg)
    end


    local function on_join(info)
        local network = hexchat.get_info("network")
        local conf = nicks[network]

        if conf ~= nil then
                local nick = info[1]
                local chan = info[2]

                if conf[chan] == true then
                        spam(nick)
                elseif type(conf[chan]) == "table" and
conf[chan][nick] == true then
                        spam(nick)
                end
        end

        return hexchat.EAT_NONE
    end


    hexchat.hook_print("Join", on_join)

You'll want to edit the configuration. The example config here
messages everyone on freenode who joins the ##test4910 channel, and
only certain nicks that join ##test4911.

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mjh215 · 2019-06-15 16:58

Thanks for the script, and also thanks for mentioning Memoserv as my
server does use that.