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Javascript 快报;无法获取/查询";错误_Javascript_Node.js_Express_Routes - Fatal编程技术网

Javascript 快报;无法获取/查询";错误

Javascript 快报;无法获取/查询";错误,javascript,node.js,express,routes,Javascript,Node.js,Express,Routes,我最近尝试使用Express,我发现这有点困难,我尝试在app.js文件和require到index.js之后定义路由,但当我尝试浏览localhost:3000/route query.js文件 我在app.js中尝试了这个 app.get('/query',require('./routes/query.js')); var queryX = require('./query.js'); app.get('/query',queryX.show); 在index.js中 app.get(

我最近尝试使用Express,我发现这有点困难,我尝试在
app.js
文件和
require
index.js
之后定义路由,但当我尝试浏览
localhost:3000/route

query.js文件

我在app.js中尝试了这个

app.get('/query',require('./routes/query.js'));
var queryX = require('./query.js');
app.get('/query',queryX.show);
index.js中

app.get('/query',require('./routes/query.js'));
var queryX = require('./query.js');
app.get('/query',queryX.show);
我在github上尝试了
路由分离
的示例,但也遇到了一个错误

为什么我不能让它工作?

仔细查看示例:和

app.js:

var site = require('./routes/site.js');
app.get('/', site.index);
var express = require('../..')
  , app = express();

require('./routes')(app);
routes/site.js:

module.exports = function(req, res) { ... };
如果要使用routes/index.js存储所有路由,则必须将
app
传递给导出的函数

比如:

app.js:

var site = require('./routes/site.js');
app.get('/', site.index);
var express = require('../..')
  , app = express();

require('./routes')(app);
routes/index.js:

var more_routes = require('./more_routes');

module.exports = function(app) {

  app.get('/', function(req, res){...});
  app.get('/show', more_routes.show);
  app.get('/list', more_routes.list);
}

谢谢,我使用了第二个例子,它起作用了,看起来这些例子没有提到一些信息或者已经过时了。