Node JS Express app.get() script in different page-CodePudding

I have the following script in my index.js page -

app.get('/', function(request, response) {
    response.render('index', {
        foo: "bar"
    })
})

This is working fine, but I want to have the response.render code in a different file, as there is going to be a lot of code behind it. Or basically, I want to get another file to call the app.get() method for the root ("/") path, and some other file to call the method for the '/admin' path, and another for '/post' path. Any way to do it ?

CodePudding user response：

What you could do is create a new javascript file with export modules to handle the routes. I'm not sure if you will be able to move the app.get() methods to a different JS file, but here's something you can do:

server.js:

const manageIndexRoute = require("./manageindex.js");
app.get('/', manageIndexRoute)

manageindex.js

module.exports = (req, res) => {
  res.render('index', {
    foo: "bar",
    ...
  })
  ...
}

CodePudding user response：

This can be achieved suing a router

You can make a new file called indexRoute and require it into your index.js

indexRoute = require("./indexRoute")

Then underneath that, add

app.use('/', indexRoute)

Now go into the indexRoute.js file and add the following code:

var express = require('express')
var router = express.Router()
 
router.get('/', function (req, res) {
  res.send('Birds home page')
})
 
module.exports = router

There. this directs all trafic from http://example.com/ to this file

But there is an alternative i guess

You can make a new file called indexHandler and require it into your index.js

indexHandler = require("./indexHandler")

Then change your app.get call to this:

app.get('/', function(request, response) => indexHandler)

Now go into the indexRoute.js file and add the following code:

indexHandler = (request, response) => {
    response.render('index', {
         foo: "bar"
    })
}

module.exports = indexHandler