section 20

dynamic route

req.params
uuid

app.get("/restaurants/:id", (req, res) => {
	const id = req.params.id;
	res.render("restaurant-detail", { id });
});

/ 를 붙여야 절대경로(도메인 뒤에 붙는다)

400 유저프론트 문제
500 서버문제. 예를 들어 많은 유저가 동시에, 파일이 없거나.

catch all error

• middleware executes on multiple incoming requests

맨 마지막줄에

• app.use()
• res.status()

app.use((req, res) => {
  res.status(404).render("404");
});

app.use((error, req, res, next) => {
  res.status(500).render("500");
});

app.listen(3000);

refactoring app.js

app.js가 너무 길고 복잡해졌다. 어떻게 간단하게 바꿀수 있을까

1. common util 분리하기

split common functions into util folder

import

const { getStoredRestaurants } = require('./util/restaurant-data');

export

const filePath = path.join(__dirname, "..", "data", "restaurants.json");

function getStoredRestaurants() {
  ///
}

module.exports = {
  getStoredRestaurants: getStoredRestaurants,
  storeRestaurants: storeRestaurants,
};

---

2. express router 분리하기

don't make new app by express.
use express.Router() as app

1) export routes

routes/default.js
express.Router()

const express = require('express');
const router = express.Router();

// app.get => router.get
router.get("/", (req, res) => {
  res.render("index");
});

module.exports = router;

2) import routes

use middleware
every request starts with / go into defaultRoutes and find (not exact path)

app.use("/", defaultRoutes);
app.use("/", restaurantRoutes);

sort

sort((a, b) => if(a > b))

a > b return 1 // flip
else return -1

query string

?...
라우팅을 결정할때 쿼리스트링은 ignored