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

如果你想把路由放在一个单独的文件中,例如routes.js,你可以这样创建routes.js文件:

module.exports = function(app){

    app.get('/login', function(req, res){
        res.render('login', {
            title: 'Express Login'
        });
    });

    //other routes..
}

然后你可以从app.js中以这样的方式传递app对象:

require('./routes')(app);

看看这些例子:https://github.com/visionmedia/express/tree/master/examples/route-separation

其他回答

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'}

如果你想要一个单独的.js文件来更好地组织你的路由,只需在app.js文件中创建一个变量,指向它在文件系统中的位置:

var wf = require(./routes/wf);

然后,

app.get('/wf', wf.foo );

其中.foo是wf.js文件中声明的某个函数。如

// wf.js file 
exports.foo = function(req,res){

          console.log(` request object is ${req}, response object is ${res} `);

}

/routes文件夹中所有.js文件的完全递归路由,把它放在app.js中。

// Initialize ALL routes including subfolders
var fs = require('fs');
var path = require('path');

function recursiveRoutes(folderName) {
    fs.readdirSync(folderName).forEach(function(file) {

        var fullName = path.join(folderName, file);
        var stat = fs.lstatSync(fullName);

        if (stat.isDirectory()) {
            recursiveRoutes(fullName);
        } else if (file.toLowerCase().indexOf('.js')) {
            require('./' + fullName)(app);
            console.log("require('" + fullName + "')");
        }
    });
}
recursiveRoutes('routes'); // Initialize it

在/routes中你放入whatevername.js并像这样初始化你的路由:

module.exports = function(app) {
    app.get('/', function(req, res) {
        res.render('index', { title: 'index' });
    });

    app.get('/contactus', function(req, res) {
        res.render('contactus', { title: 'contactus' });
    });
}

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

我试图用“express”更新这个答案:“^4.16.3”。这个答案与ShortRound1911的答案相似。

server.js:

const express = require('express');
const mongoose = require('mongoose');
const bodyParser = require('body-parser');
const db = require('./src/config/db');
const routes = require('./src/routes');
const port = 3001;

const app = new express();

//...use body-parser
app.use(bodyParser.urlencoded({ extended: true }));

//...fire connection
mongoose.connect(db.url, (err, database) => {
  if (err) return console.log(err);

  //...fire the routes
  app.use('/', routes);

  app.listen(port, () => {
    console.log('we are live on ' + port);
  });
});

/ src /线路/ index.js:

const express = require('express');
const app = express();

const siswaRoute = require('./siswa_route');

app.get('/', (req, res) => {
  res.json({item: 'Welcome ini separated page...'});
})
.use('/siswa', siswaRoute);

module.exports = app;

/ src /线路/ siswa_route.js:

const express = require('express');
const app = express();

app.get('/', (req, res) => {
  res.json({item: 'Siswa page...'});
});

module.exports = app;