In my Express application I implement routes in routes.ts:
var Router = express.Router();
Router.route('/models/:modelId')
.get(function (req, res) {
service.setParameter(req)
service.get(req,res)
});
Router.route('/models/:modelId')
.post(function (req, res) {
service.setParameter(req)
service.post(req,res)
});
And express.ts:
export const App = express()
App.use(express.json())
App.use(express.urlencoded({ extended: true }))
App.use(helmet())
App.use('/', Router)
At each router call I'd like to execute a piece of code service.setParameter(req) that gets particular parameter from 'params', but I don't want to add to each router method explicitly.
I tried adding it at as middleware before and after Router
App.use('/', Router)
App.use(function(req, res, next){
service.setParameter(req)
next()
})
But if I define it before Router then route hasn't been set yet, and I don't get the parameter I want, and if I define it after, then middleware is not executed.
How can execute service.setParameter(req) in a generic way so that it applies to all the routes?
In express.ts file, you can add a middleware that would do it before mounding the Router, and then just procced forward with next(). You can do it like this:
App.use('/*', (req, res, next) => {
service.setParameter(req);
next();
});
App.use('/', Router)
You need to place your custom middleware between the context path and your router inside app.use(..):
const router = express.Router();
router.post('/', (req, res) => {
service.post(req,res);
});
router.get('/', (req, res) => {
service.get(req,res)
});
app.use('/models', (req, res, next) => {
service.setParameter(req);
next();
}, router);
With above code the middleware will be excecuted for all requests to '/models'.
You can use app.use(async (req,res,next) => {...}) in order to declare a middleware that executes in all the requests, if you want this middleware to be called first, it must be declare before than your routes, the middleware have to call next() in order to continue with the execution flow, if you want to be called at the end of you request, you have to put at the end of your declarations but before of the error middleware, in that approach each route have to call next() at the end of your function.
First approach
const express = require('express');
const app = express();
const router = express.Router();
router.post('/', async (req, res) => {
await service.post(req,res);
});
router.get('/', async (req, res) => {
await service.get(req,res)
});
app.use((req,res,next) => {
console.log("always called");
next();
});
app.use('/',router);
Second approach
const express = require('express');
const app = express();
const router = express.Router();
router.post('/', async (req, res, next) => {
await service.post(req,res);
next();
});
router.get('/', async (req, res, next) => {
await service.get(req,res);
next();
});
app.use('/',router);
app.use((req,res) => {
console.log("always called");
});
Thanks for all the answers, they helped me better understand how routing works on Express.
I found another solution, which I think works best in my case - using Router.all() method:
const setRequest = function(req, res, next){
logger.setRequest(request)
next()
}
Router.route('/models/:model_id')
.all(setRequest)
.get(function (req, res) {service.execute()})
.put(function (req, res) {service.execute()})
Related
app.js:
var app = express();
app.use('/my-page', require('./routes/my-page.js'));
my-page.js:
const router = require('express').Router();
router.get('/one', function (req, res, next) {
return res.send('this is /my-page/one');
});
router.get('/my-other-page', function (req, res, next) {
return res.send('this is /my-other-page');
});
How do I make it so my-other-page isn't under my-page, but is instead on the root? I do not want to change app.js because i still want most routes under that page, just one specific one that I want to not have /my-page.
I tried .. in the route but doesnt work. I tried making app from app.js global, but that didn't seem to work either.
Typically you have a routes.js that can import the other routes. (You don't need to, but I think it will better for your structure).
app.js
var app = express();
app.use('/', require('./routes/routes.js'));
routes.js
const router = require('express').Router();
router.use('/my-page', require('./my-page.js'));
router.use('/my-other-page', require('./my-other-page.js'));
Then you split my-page.js and my-other-page.js.
my-page.js
const router = require('express').Router();
// This is /my-page/one
router.get('/one', function (req, res, next) {
return res.send('this is /my-page/one');
});
my-other-page.js
// This is /my-other-page/one
router.get('/one', function (req, res, next) {
return res.send('this is /my-other-page/one');
});
app.js
// Calling Routes
require("./routes")(app);
router folder
index.js
module.exports = function (app) {
app.use("/", require("./all_routes"));
}
all_routes.js
var express = require("express");
var router = express.Router();
router.get("/", function (req, res, next) {
res.render("home/index.html");
});
//About Page
router.get("/about", function (req, res, next) {
res.render("about/index.html");
});
//Contact
router.get("/contact", function (req, res, next) {
res.render("contact/index.html");
});
//product
router.get("/product", function (req, res, next) {
res.render("product/index.html");
});
//product list
router.get("/product/demo-product", function (req, res, next) {
res.render("demo-product/index.html");
});
router.get("/product/request-product", function (req, res, next) {
res.render("request-product/index.html");
});
//service
router.get("/service", function (req, res, next) {
res.render("product/index.html");
});
//service list
router.get("/service/what-we-do", function (req, res, next) {
res.render("what-we-do/index.html");
});
router.get("/service/how-we-do", function (req, res, next) {
res.render("how-we-do/index.html");
});
I am trying to reduce the code in all_routes.js file has same code is repeating again and again
I searched online and trying to create it dynamically but getting no success is there any way I can reduce the line of code as I have given the follow of my code above
If you'd like to cut down on boilerplate of all your get routes, one option is to create an object to map your routes to the files they're loading. Then you can iterate over that object and add the routes to your router.
const routes = {
"/": "home/index.html",
"/about": "about/index.html",
"/contact": "contact/index.html"
// Add others here
}
for (let key in routes) {
router.get(key, function (req, res, next) {
res.render(routes[key]);
});
}
Edit: If your routes are consistent in that the index.html file will always be in the directory named after the part after the last / in your route, you can potentially use an array and some fancy logic. Just don't break the rule!
const routes = [
"/contact",
"/product",
"/product/demo-product",
"/product/request-product"
]
routes.forEach(route => {
const path = /[^/]*$/.exec(route)[0];
router.get(route, function (req, res, next) {
res.render(`${path}/index.html`);
});
})
How can I insert isLoggedIn as a condition to the get request using router.route?
const controller = require('./controller');
const Router = require('express').Router;
const router = new Router();
function isLoggedIn(req, res, next) {
if (req.isAuthenticated())
return next();
res.redirect('/');
}
router.route('/')
.get((...args) => controller.find(...args))
I assume that the ...args are (req, res, next)
I tried
router.route('/')
.get(isLoggedIn(...args) => controller.find(...args))
But I get
.get((isLoggedIn(...args)) => controller.find(...args))
^
SyntaxError: Unexpected token (
The docs say, that you can assign multiple handlers to one route. Like this:
app.use('/user/:id', function (req, res, next) {
console.log('Request URL:', req.originalUrl)
next()
}, function (req, res, next) {
console.log('Request Type:', req.method)
next()
})
Source
In your case the coding looks like the following
router.get('/', isLoggedIn, controller.find);
so I have a route written with app.route:
app.route('/word/:id')
.get(word.getWord)
.put(word.updateWord)
.delete(word.deleteWord);
Now I want to re-write the route with some middleware. I know how to do this route by route, for example the first route would be:
app.get('/word/:id', connect.ensureLoggedIn('/auth/facebook'), word.getWord);
But can I apply the same middleware to all 3 routes at once?
you can use something like
var express = require('express');
var router = express.Router();
// -- #section router middleware
router.use(function (req, res, next) {
/* your code here */
next();
});
// -- #section routes
router.get('/users', function (req, res) {
//res.render('users')
});
router.post('/users', function (req, res) {
});
The below code not functioning as expected
var express = require('express');
var app = express();
app.use(function(req, res, next) {
console.log('first text');
next();
}, function (req, res, next) {
console.log('secondText');
res.end()
}).listen(3000)
app.use([path,] function [, function...])
Mounts the specified middleware function or functions at the specified path. If path is not specified, it defaults to '/' in express documentation but I can't run the second function, not sure why. When I try localhost:3000 in Firefox I receive Cannot GET /
This code is working, but previously it wasn't working:
var express = require('express');
var app = express();
app.use(function(req, res, next) {
console.log("one");
next();
})
.use(function(req,res,next){
console.log("second");
res.end()
})
.listen(3000)
Could it be because of the missing .?
var express = require('express');
var app = express();
app.use(function(req, res, next) {
console.log('first text');
next();
}, function (req, res, next) {
console.log('secondText');
}).listen(3000) //Was missing a period