how to create a nodeJS module with expressJS - node.js

Question

I'm making a nodeJS module, and I want to use expressJS as a framework for it.
I'm trying to see, how I could go by, including a function inside and app.get(); and call it via another file, such as the actual app.
var express = require("express");
var app = express();
app.get("/", function (req, res) {
exports.type = function (text) {
console.log(req.ip);
console.log(text);
}
});
now when I use this, and i call it on the actual app like:
var web = require("directory_to_file");
var express = require("express");
var app = express();
var http = require("http").Server(app);
app.get("/", function (req, res) {
web.type("Hello, world");
});
http.listen(10022, function () {
console.log("server is up");
});
I get an error:
TypeError: Property 'type' of object #<Object> is not a function
anyone know a way to make it so I can call the function?

Answer 1

There are generally two things you want to export as a module - an API and a Middleware. The classic example of middleware is an authentication module. To do the middleware, just export the middleware. I tend to do a little more than that so I can configure the middleware later. Something along the lines of this:
module.exports = exports = function(config) {
// Do something with config here
return function(req, res, next) {
// your middleware here
};
};
You can then use your middleware in your main program like this:
var app = require('express')(),
mymodule = require('./mymodule');
var config = {}; // replace with whatever config you need
app.use(mymodule(config));
app.listen(process.env.PORT || 3000);
To implement an API, you will create a Router object, then attach your routes to the Router object. You can then "mount" your router in your main program. For example, you could have a file called 'myroutes.js' with the following contents:
var express = require('express'),
myroutes = express.Router();
myroutes.get('/foo', (req, res) => {
res.status(200).type('application/json').send({ myparam: 'foo' });
});
module.exports = exports = myroutes;
Have the following in your main program:
var app = require('express')(),
myroutes = require('./myroutes');
app.use('/api', require('./myroutes'));
app.listen(process.env.PORT || 3000);
Here, in 'myroutes.js', I'm defining a sub-route of /foo and then in the main program, I'm mounting that on /api - so I would access /api/foo to access that API.

Answer 2

In your directory_to_file you are only exporting on app.get('/') which will never be called.
You could add in your directory_to_file the following code
var express = require('express');
var router = express.Router();
router.get('/', function(req, server) {
console.log(req.ip);
});
module.exports = router;
And in your main file you could use app.use('/', web)
A short explanation:
You are creating a new express app / config in your directory_to_file file which won't be launched or used. So your app.get event won't be fired once.
That's why web.type is not a function. You are not exporting anything.
Use the way I provided. This is a commonly used method.
You could call the code I provided a "route". Create multiple routes / route files and include them in your main method.

Answer 3

Your code just looks confused. If I understand you correctly, what you are really trying to do (at least in Node/express terminology) is write your own middleware.
Express is designed with this in mind and it's pretty straightforward e.g.
ipLogger.js
module.exports = function(req, res, next) {
console.log(req.ip);
next();
}
app.js
var http = require("http")
, express = require("express");
, app = express()
, server = http.Server(app)
, ipLogger = require("./ipLogger.js");
app.use(ipLogger()); // log IP of all requests
// handle routes
server.listen(10022, function() {
console.log("server is up");
});