Express:单独的路由和控制器文件

min*_*rse 7 node.js express

我试图在Express中将一些路由及其处理程序逻辑拆分为单独的文件.我已经看到了类似于JS的示例目录结构,其中使用了单独的路由和控制器文件,因此我试图实现这种方法但是遇到了问题.

我的服务器和路由配置如下:

server.js

var express = require('express'),
  app = express(),
  bodyParser = require('body-parser'),
  routes = require('./routes/index')

app.use(bodyParser.urlencoded({extended: true}));
app.use(bodyParser.json());
routes(app);
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/routes/index.js

module.exports = function(app) {
  var catalogues = require('../routes/catalogues');
  app.use('/catalogues-api', catalogues);
};
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/routes/catalogues.js

var catalogues = require('../controllers/catalogues');
module.exports = function(app) {

  app.route('/catalogues')
    .get(catalogues.apiGET)
    .post(catalogues.apiPOST);
};
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/controllers/catalogues.js

var request = require('request');

exports.apiGET = function(req, res) {
  var options = prepareCataloguesAPIHeaders(req);
  request(options, function(err, response, body){
    res.send(body);
  });
};

exports.apiPOST = function(req, res) {
  var options = prepareCataloguesAPIHeaders(req);
  options.json = true;
  options.body = stripBody(req.body);
  request(options, function(err, response, body){
    res.send(body);
  });
};
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运行应用程序并对/ catalogs-api/catalogs发出GET请求时,我从节点抛出一个错误:

TypeError:undefined不是module.exports的函数(C:\ Users\rparker\Documents\GitHub\testproj\src\server\routes\catalogues.js:4:7)

这是在我的/routes/catalogues.js文件中引用app.route声明.我在设置中显然遗漏了一些东西,但我无法理解.

有人可以帮忙吗?谢谢

ma0*_*a08 9

/routes/catalogues.js

var express = require('express');
var router = express.Router();
var catalogues = require('../controllers/catalogues');

router.route('/catalogues')
.get(catalogues.apiGET)
.post(catalogues.apiPOST);
module.exports = router;
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