Re: Using mod_lua to modify request body before being sent to mod_proxy

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Yeah, I misspoke when I said LuaHookInsertFilter; that isn’t actually implemented yet at least on 2.4. Actually have been trying the quick handler and hookfixups.

What hook should I be using? What value do I return from my handler so that Apache will still send the request onto the reverse proxy?

I feel like this (http://www.modlua.org/recipes/loadbalancing) may be what I want, but I don’t know how to hook that into my httpd.conf file. Does this tie into mod_proxy at all, or is this instead of?

In that example, what is the magic string “proxy-server”?:

function proxy_handler(r)
    r.handler  = "proxy-server"
    r.proxyreq = apache2.PROXYREQ_REVERSE
    r.filename = "proxy:" .. backends[math.random(1,#backends)] .. r.uri
    return apache2.DECLINED -- let the proxy handler do this instead
end





On April 15, 2016 at 9:47:01 PM, Eric Covener (covener@xxxxxxxxx) wrote:

On Fri, Apr 15, 2016 at 9:10 PM, Matt Hughes <hughes.matt@xxxxxxxxx> wrote:
> I currently have Apache setup as a reverse proxy to another server.
>
> I was hoping to use mod_lua to slightly tweak the request body of incoming
> requests before sending them off to the other server. I have mod_lua setup
> enough so that I have a LuaHookInsertFilter that is pointing to my
> LuaInputFilter. But the question is, what does the function in the
> LuaHookInsertFilter have to return so that the modified request continues on
> to the proxy?
>
> Right now I have something like:
>
> function handle(r)
> r.info(r, "example handler called") -- debug print
> r:add_input_filter("myInputFilter")
> return apache2.PROXYREQ_REVERSE
> end
>
> Am I using the right LuaHook? Any examples would be greatly appreciated.

Looks right to me. The underlying hook for "insert_filter" expects a
void return.

This means everyone who is registered gets called every time.

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