我正在尝试为localhost创建路由:port / admin / 我想将routes.js文件和view.js文件保存在匹配的文件夹中,以便我以后不会有太多的spaggeti
但我一直得到:500错误:无法查找视图“/ admin / manage_subjects” 尝试创建新路由并使用相同的文件夹
我有以下带快递的视图文件夹
mainapp(root)
routes(folder)
admin(folder)
index.js(nested inside admin)
index.js(nested inside routes)
views(folder)
admin(folder)
admin_layout.jade(nested inside admin)
manage_subjects.jade(nested inside admin)
index.jade(nested inside views)
layout.jade(nested inside views)
代码: 路由/管理/ index.js
exports.index = function (req, res) {
res.render('manage_subjects',{title:'Express'});}
视图/管理/ manage_subjects.jade
extends admin_layout
block content
h1 = title
p Welcome to #{title}
我的app.js代码
/**
* Module dependencies.
*/
var express = require('express')
, routes = require('./routes')
, admin_routes = require('./routes/admin/')
, user = require('./routes/user')
, http = require('http')
, path = require('path')
, repository = new (require('./domain_model/repository'))();
var app = express();
// all environments
app.set('port', process.env.PORT || 3000);
app.set('views', __dirname + '/views');
app.set('view engine', 'jade');
app.use(express.favicon());
app.use(express.logger('dev'));
app.use(express.bodyParser());
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());
}
//fill local subjects
repository.subjects.GetAll(function (err, data) {
if (err) throw err;
app.locals.subjects = data;
});
//append routes
app.get('/', routes.index);
app.get('/admin', admin_routes.index);
app.get('/users', user.list);
http.createServer(app).listen(app.get('port'), function(){
console.log('Express server listening on http://localhost:' + app.get('port'));
});
答案 0 :(得分:14)
我一直在处理我认为同样的问题,并想出如何解决它。因此,如果有其他人遇到这个问题我会发布我的解决方案。
所以这就是我所拥有的导致404和500的
app.js
var routes = require('./routes/index');
var admin = require('./routes/admin');
app.use('/', routes);
app.use('/admin', admin);
这是我的routes / index.js
//append routes
var express = require('express');
var router = express.Router();
router.get('/', function(req, res) {
res.render('index', {title: 'Express'});
});
module.exports = router;
和我的路线/ admin.js:
var express = require('express');
var router = express.Router();
router.get('/admin', function(req, res) {
res.render('admin/index', {title: 'Express'});
});
module.exports = router;
通过在router.get()函数中定义第二个/ admin我认为我有效地告诉节点在我的views文件夹中查找以下路径视图/ admin / admin / index.ejs中的html。所以要解决这个问题,我要做的就是从router.get()中删除/ admin,或者从app.use()中删除/ admin 所以我的工作代码现在看起来像这样:
app.js
var routes = require('./routes/index');
var admin = require('./routes/admin');
app.use('/', routes);
app.use('/admin', admin); //I left the /admin here and instead removed the one in routes/admin.js
这是我的routes / index.js
//append routes
var express = require('express');
var router = express.Router();
router.get('/', function(req, res) {
res.render('index', {title: 'Express'});
});
module.exports = router;
和我的路线/ admin.js:
var express = require('express');
var router = express.Router();
router.get('/', function(req, res) { //NOTICE THE CHANGE HERE
res.render('admin/index', {title: 'Express'});
});
module.exports = router;
因此,进行更改后,我的视图文件夹中可以包含子文件夹。
答案 1 :(得分:1)
我会查看TJ关于模块化应用on his vimeo
的视频,关于此工作流程的最佳部分是您的代码变得非常灵活,并且更容易留下DRY
。
此外,我会使用app.set("views")
var path = require("path");
app.set('views', path.join(__dirname, 'views'));
// you can then extend this to the example for routes
另一个替代方案就像在app.js
文件中一样:
var express require("express")
var app = express()
var routes = require("./path/to/routes")(app)
然后路线看起来像:
routes = function (app) {
app.get("/route", middleWareifYou.gotIt, route.handler || function (req, res) {
res.send("some msg");
});
};
module.exports = routes
干杯,希望这有帮助!
答案 2 :(得分:0)
简单答案,用于views文件夹中的子文件夹(我的称为前端)
click here the picture to see the folder structure
$ readelf --string-dump=.gnu_debuglink helloworld | sed -n '/]/{s/.* //;p;q}'
helloworld