在我的NodeJS express应用程序中,我有app.js,它有一些常见的路由。然后在wf.js文件中,我想定义更多的路由。

如何让app.js识别在wf.js文件中定义的其他路由处理程序?

一个简单的要求似乎不起作用。


当前回答

index.js

const express = require("express");
const app = express();
const http = require('http');
const server = http.createServer(app).listen(3000);
const router = (global.router = (express.Router()));
app.use('/books', require('./routes/books'))
app.use('/users', require('./routes/users'))
app.use(router);

路线/ users.js

const router = global.router
router.get('/', (req, res) => {
    res.jsonp({name: 'John Smith'})
}

module.exports = router

路线/ books.js

const router = global.router
router.get('/', (req, res) => {
    res.jsonp({name: 'Dreams from My Father by Barack Obama'})
}

module.exports = router

如果您的服务器在本地运行(http://localhost:3000),那么

// Users
curl --request GET 'localhost:3000/users' => {name: 'John Smith'}

// Books
curl --request GET 'localhost:3000/books' => {name: 'Dreams from My Father by Barack Obama'}

其他回答

您可以将所有路由函数放在其他文件(模块)中,并将其链接到主服务器文件。 在main express文件中,添加一个将模块链接到服务器的函数:

   function link_routes(app, route_collection){
       route_collection['get'].forEach(route => app.get(route.path, route.func));
       route_collection['post'].forEach(route => app.post(route.path, route.func));
       route_collection['delete'].forEach(route => app.delete(route.path, route.func));
       route_collection['put'].forEach(route => app.put(route.path, route.func));
   }

并为每个路由模型调用该函数:

link_routes(app, require('./login.js'))

在模块文件(例如- login.js文件)中,像往常一样定义函数:

const login_screen = (req, res) => {
    res.sendFile(`${__dirname}/pages/login.html`);
};

const forgot_password = (req, res) => {
    console.log('we will reset the password here')
}

然后用request方法导出它作为键,值是一个对象数组,每个对象都有路径和功能键。

module.exports = {
   get: [{path:'/',func:login_screen}, {...} ],
   post: [{path:'/login:forgotPassword', func:forgot_password}]
};   

如果你使用express-4。使用TypeScript和ES6,这将是最好的模板使用:

登录src /火/ ts。

import express, { Router, Request, Response } from "express";

const router: Router = express.Router();
// POST /user/signin
router.post('/signin', async (req: Request, res: Response) => {
    try {
        res.send('OK');
    } catch (e) {
        res.status(500).send(e.toString());
    }
});

export default router;

src / app.ts

import express, { Request, Response } from "express";
import compression from "compression";  // compresses requests
import expressValidator from "express-validator";
import bodyParser from "body-parser";
import login from './api/login';

const app = express();

app.use(compression());
app.use(bodyParser.json());
app.use(bodyParser.urlencoded({ extended: true }));
app.use(expressValidator());

app.get('/public/hc', (req: Request, res: Response) => {
  res.send('OK');
});

app.use('/user', login);

app.listen(8080, () => {
    console.log("Press CTRL-C to stop\n");
});

比使用var和module.exports干净多了。

index.js

const express = require("express");
const app = express();
const http = require('http');
const server = http.createServer(app).listen(3000);
const router = (global.router = (express.Router()));
app.use('/books', require('./routes/books'))
app.use('/users', require('./routes/users'))
app.use(router);

路线/ users.js

const router = global.router
router.get('/', (req, res) => {
    res.jsonp({name: 'John Smith'})
}

module.exports = router

路线/ books.js

const router = global.router
router.get('/', (req, res) => {
    res.jsonp({name: 'Dreams from My Father by Barack Obama'})
}

module.exports = router

如果您的服务器在本地运行(http://localhost:3000),那么

// Users
curl --request GET 'localhost:3000/users' => {name: 'John Smith'}

// Books
curl --request GET 'localhost:3000/books' => {name: 'Dreams from My Father by Barack Obama'}

我写了一个小插件来做这件事!我厌倦了一遍又一遍地写同样的代码。

https://www.npmjs.com/package/js-file-req

希望能有所帮助。

所有这些答案的一个调整:

var routes = fs.readdirSync('routes')
      .filter(function(v){
         return (/.js$/).test(v);
      });

只需使用正则表达式通过测试数组中的每个文件进行筛选。它不是递归的,但它会过滤掉不以.js结尾的文件夹