Express js：将多个路由文件转换为单个文件

Question

在我的express.js应用程序中，我按照以下方式组织路线

路由/comment.js（处理所有评论路由）

var express = require('express');
var router = express.Router();
var comment = require('../controller/comment');
router.route('/new').get(comment.create);
module.exports = router;

route / post.js（处理所有发布路线）

var express = require('express');
var router = express.Router();
var post = require('../controller/post');
router.route('/new').get(post.create);
module.exports = router;

我以这种方式将其包含在app.js文件中

//need to include these declarations into another file and include that file here
  //eg: require('config/main-routes');

    /*
     routes like comment/new
    */
    var comment  = require('./routes/comment');
    app.use('/comment',comment);
    /*
     routes like post/new
    */
    var post  = require('./routes/post');
    app.use('/post',post)

这工作正常，但是我想将此包含到另一个文件中，然后说config/route-main.js并将其链接到app.js文件

我怎样才能做到这一点？？

Answer 1

在app.js包括

var routes  = require('./config/route-main');
app.use('/',routes);

并在route-config.js中

var express = require('express');
var router = express.Router();
var comment  = require('./routes/comment');
var post  = require('./routes/post');

router.use('/comment',comment);
router.use('/post',post)


module.exports = router

检查此链接http://katieleonard.ca/blog/2016/nested-routes-with-expressjs/