Node.js 将app.get()放置在单独的文件中
我有一个普通的expressjs app.jsNode.js 将app.get()放置在单独的文件中,node.js,express,Node.js,Express,我有一个普通的expressjs app.js /** * Module dependencies. */ var express = require('express'); var routes = require('./routes'); var user = require('./routes/user'); var http = require('http'); var path = require('path'); var app = express(); // all en
/**
* Module dependencies.
*/
var express = require('express');
var routes = require('./routes');
var user = require('./routes/user');
var http = require('http');
var path = require('path');
var app = express();
// all environments
app.set('port', process.env.PORT || 3000);
app.set('views', path.join(__dirname, 'views'));
app.set('view engine', 'jade');
app.use(express.favicon());
app.use(express.logger('dev'));
app.use(express.json());
app.use(express.urlencoded());
app.use(express.methodOverride());
app.use(app.router);
app.use(express.static(path.join(__dirname, 'public')));
// development only
if ('development' == app.get('env')) {
app.use(express.errorHandler());
}
app.get('/', routes.index);
app.get('/users', user.list);
http.createServer(app).listen(app.get('port'), function(){
console.log('Express server listening on port ' + app.get('port'));
});
我想把它去掉
app.get('/', routes.index);
app.get('/users', user.list);
把它们放在一个单独的文件里。我正在尝试这个
module.exports = function(/* any dependency? */){
app.get('/', routes.index);
app.get('/users', user.list);
}
在app.js
文件中,我有require('./routed.js')代码>
但这不起作用。我应该如何解决这个问题?试试这个:
app.use(express.static(path.join(__dirname, 'public')));
require('./routed.js')(app);
在/routed.js中
module.exports = function(app){
app.get('/', routes.index);
app.get('/users', user.list);
}
var routes = require('./routes');
var user = require('./routes/user');
module.exports = function(app){
app.get('/', routes.index);
app.get('/users', user.list);
}
require('./routed.js')(app);
我接受了克拉苏的回答,稍作修改
routed.js
module.exports = function(app){
app.get('/', routes.index);
app.get('/users', user.list);
}
var routes = require('./routes');
var user = require('./routes/user');
module.exports = function(app){
app.get('/', routes.index);
app.get('/users', user.list);
}
require('./routed.js')(app);
在app.js的
module.exports = function(app){
app.get('/', routes.index);
app.get('/users', user.list);
}
var routes = require('./routes');
var user = require('./routes/user');
module.exports = function(app){
app.get('/', routes.index);
app.get('/users', user.list);
}
require('./routed.js')(app);
这是可行的。我对它进行了一些修复,但它可以工作,但我不知道它带来的性能成本。您需要var routes=require('./routes');var user=require('./路由/用户')在routed.js
文件中的code>,以使其正常工作。