next('route') not working !

[linqFR]

I am using a router (express.Router - called testrouter in the code here) added to main app (express) and i add subrouter (express.Router) to this router/testrouter with a function just before it to test if request is from specific referer, but the subrouter is never skipped!

const express = require('express');
const http = require('http');
const httpport = process.env.PORT || 80; //http

app = express();

var testrouter = express.Router();
app.use(testrouter);

var subrouter = express.Router();

var subrouterRoute = '/blabla'
testrouter.use(subrouterRoute, 
    function(req, res, next){
        console.log('trying to go into subrouter');
        if (req.headers.referer) {console.log('going', typeof req.headers.referer), next();}
        else {console.log('not going', typeof req.headers.referer); next('route');}
    },
    subrouter,
    function(req,res,next){
        console.log('gone out of subrouter');
        next();
    }
);

subrouter.use(function(req,res,next){
    console.log('IN SUBROUTER');
    next();
})
http
.createServer(app)
.listen(httpport, function(){
    console.log("server http listened @ : ", this.address());
});

if i request /blabla directly, there is no referer so it is undefined... but though, the console logs the following:

trying to go into subrouter
not going undefined (this is the referer)
IN SUBROUTER
gone out of subrouter

[tunniclm]

The next('route') facility doesn't skip middleware added through the use() function, see: http://expressjs.com/en/guide/using-middleware.html

Excerpt:

To skip the rest of the middleware functions from a router middleware stack, call next('route') to pass control to the next route. NOTE: next('route') will work only in middleware functions that were loaded by using the app.METHOD() or router.METHOD() functions.

If you change:

subrouter.use(function(req,res,next){
    console.log('IN SUBROUTER');
    next();
})

to:

subrouter.get(function(req,res,next){
    console.log('IN SUBROUTER');
    next();
})

...it should work.

[linqFR]

ok a bit tricky... use not included then !
and that requires to write code for every case !!!

[tunniclm]

@linqOxelar all() also works if you need to match all methods.

[linqFR]

@tunniclm thanks for hint!