Express js根据路线需要不同的模块

Question

I have an app I am trying to mount in express using app.use , such as:

app.use(require('./foo'));

This will return the app in the sibling file foo.js when defined as such:

var router = require('express').Router({ mergeParams: true });
module.exports = router;
router.get('/', function (req, res) { 
   res.send(200);
}

// Navigate to /foo  ->  200 ok! 

However, I would like to reference a different app based on the request path, so that instead of

app.use(require('./foo'));

We could do

app.use(function (req, res) {
    require(req.path);
};

So if there was some file bar.js , we could conceivably

// Navigate to /bar 
returns bar app. 

But this does not work, and instead times out, as I believe it is not mounting the required app correctly. How can I pass a callback to app.use that can mount a required app that is defined in the callback's function definition?

Thanks for any help.

Answer 1

A few things to consider here :

1) You will need to call the router itself (which is a (req, res) function like any middleware) and pass it your request.

app.use(function (req, res) {
    var router = require('.' + req.path); // or whatever your lookup method looks like
    router(req, res);
});

2) Since the main middleware above isn't mounted on any path, req.path will stay the same (eg '/foo' ) and will never match the child router's .get('/') route. You can work around this by rather having the child router .use() its middleware, which is fine as you are already in an application endpoint. (Note that if your middleware is simple enough you could also just export the middleware itself rather than the whole router)

// foo.js
router.use(function (req, res) {
    res.send(200);
});

Answer 2

Check out the documentation for Express req.path.

http://expressjs.com/en/api.html

It looks like req.path will return something like /foo , but you need it to be ./foo for it to find the file and require it. Try changing your require to be

require('.' + req.path);