node.js 在 Express-js 中使用路由
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Using routes in Express-js
提问by Andreas Stokholm
So I'm starting to use Node.js. I saw the video with Ryan Dahl on Nodejs.organd heard he recommended Express-js for websites.
所以我开始使用 Node.js。我在Nodejs.org上看到了 Ryan Dahl 的视频，听说他为网站推荐了 Express-js。
I downloaded the latest version of Express, and began to code. I have a fully fledged static view up on /, but as soon as I try sending parameters, I get errors like this:
我下载了最新版本的 Express，并开始编写代码。我在 / 上有一个完全成熟的静态视图，但是一旦我尝试发送参数，就会收到如下错误：
Cannot GET /wiki
I tried following the guide on expressjs.combut the way one uses routes has changed in the latest version, which makes the guide unusable.
我尝试按照expressjs.com上的指南进行操作，但是在最新版本中使用路由的方式发生了变化，这使得该指南无法使用。
Guide:
指导：
app.get('/users/:id?', function(req, res, next){
var id = req.params.id;
if (id) {
// do something
} else {
next();
}
});
Generated by Express:
由 Express 生成：
app.get('/', routes.index);
My problem arises when I try and add another route.
当我尝试添加另一条路线时，我的问题出现了。
app.get('/wiki', routes.wiki_show);
I've tried a bunch of approaches, but I keep getting the Cannot GET /wiki(404) error.
我尝试了很多方法，但我不断收到Cannot GET /wiki(404) 错误。
routes/index.js looks like this:
route/index.js 看起来像这样：
exports.index = function(req, res) {
res.render('index', { title: 'Test', articles: articles, current_article: current_article, sections: sections })
};
The only thing I did there was add some parameters (arrays in the same file) and this i working. But when I copy the contents and change exports.indexto exports.wikior exports.wiki_showI still get the Cannot GET /wikierror.
我在那里做的唯一一件事就是添加一些参数（同一文件中的数组），这就是我的工作。但是当我复制内容并更改exports.index为exports.wikior 时，exports.wiki_show我仍然收到Cannot GET /wiki错误消息。
Can anyone explain to me what I'm missing here? - Thanks.
谁能向我解释我在这里缺少什么？- 谢谢。
回答by Andreas Stokholm
So, after I created my question, I got this related list on the right with a similar issue: Organize routes in Node.js.
所以，在我创建我的问题之后，我在右边得到了一个类似问题的相关列表：在 Node.js 中组织路由。
The answer in that post linked to the Express repo on GitHuband suggests to look at the 'route-separation' example.
该帖子中的答案链接到GitHub 上的Express 存储库，并建议查看“路由分离”示例。
This helped me change my code, and I now have it working. - Thanks for your comments.
这帮助我更改了代码，现在我可以使用它了。- 感谢您的意见。
My implementation ended up looking like this;
我的实现最终看起来像这样；
I require my routes in the app.js:
我需要在 app.js 中使用我的路由：
var express = require('express')
, site = require('./site')
, wiki = require('./wiki');
And I add my routes like this:
我像这样添加我的路线：
app.get('/', site.index);
app.get('/wiki/:id', wiki.show);
app.get('/wiki/:id/edit', wiki.edit);
I have two files called wiki.js and site.js in the root of my app, containing this:
我的应用程序根目录中有两个名为 wiki.js 和 site.js 的文件，其中包含以下内容：
exports.edit = function(req, res) {
var wiki_entry = req.params.id;
res.render('wiki/edit', {
title: 'Editing Wiki',
wiki: wiki_entry
})
}
回答by systemovich
The route-mapexpress example matches url paths with objects which in turn matches http verbs with functions. This lays the routing out in a tree, which is concise and easy to read. The apps's entities are also written as objects with the functions as enclosed methods.
的路由映射快车例如匹配的URL与对象这反过来匹配HTTP动作与功能路径。这将路由布置在树中，简洁易读。应用程序的实体也被编写为对象，函数作为封闭的方法。
var express = require('../../lib/express')
, verbose = process.env.NODE_ENV != 'test'
, app = module.exports = express();
app.map = function(a, route){
route = route || '';
for (var key in a) {
switch (typeof a[key]) {
// { '/path': { ... }}
case 'object':
app.map(a[key], route + key);
break;
// get: function(){ ... }
case 'function':
if (verbose) console.log('%s %s', key, route);
app[key](route, a[key]);
break;
}
}
};
var users = {
list: function(req, res){
res.send('user list');
},
get: function(req, res){
res.send('user ' + req.params.uid);
},
del: function(req, res){
res.send('delete users');
}
};
var pets = {
list: function(req, res){
res.send('user ' + req.params.uid + '\'s pets');
},
del: function(req, res){
res.send('delete ' + req.params.uid + '\'s pet ' + req.params.pid);
}
};
app.map({
'/users': {
get: users.list,
del: users.del,
'/:uid': {
get: users.get,
'/pets': {
get: pets.list,
'/:pid': {
del: pets.del
}
}
}
}
});
app.listen(3000);
回答by Shlomi Loubaton
Seems that only index.js get loaded when you require("./routes") . I used the following code in index.js to load the rest of the routes:
似乎只有 index.js 在您 require("./routes") 时才会加载。我在 index.js 中使用了以下代码来加载其余的路由：
var fs = require('fs')
, path = require('path');
fs.readdirSync(__dirname).forEach(function(file){
var route_fname = __dirname + '/' + file;
var route_name = path.basename(route_fname, '.js');
if(route_name !== 'index' && route_name[0] !== "."){
exports[route_name] = require(route_fname)[route_name];
}
});
回答by Maleck13
You could also organise them into modules. So it would be something like.
您也可以将它们组织成模块。所以它会是这样的。
./
controllers
index.js
indexController.js
app.js
and then in the indexController.js of the controllers export your controllers.
然后在控制器的 indexController.js 中导出您的控制器。
//indexController.js
module.exports = function(){
//do some set up
var self = {
indexAction : function (req,res){
//do your thing
}
return self;
};
then in index.js of controllers dir
然后在控制器目录的 index.js 中
exports.indexController = require("./indexController");
and finally in app.js
最后在 app.js
var controllers = require("./controllers");
app.get("/",controllers.indexController().indexAction);
I think this approach allows for clearer seperation and also you can configure your controllers by passing perhaps a db connection in.
我认为这种方法可以实现更清晰的分离，并且您还可以通过传入数据库连接来配置控制器。
回答by Calvintwr
No one should ever have to keep writing app.use('/someRoute', require('someFile')) until it forms a heap of code.
没有人应该一直写 app.use('/someRoute', require('someFile')) 直到它形成一堆代码。
It just doesn't make sense at all to be spending time invoking/defining routings. Even if you do need custom control, it's probably only for some of the time, and for the most bit you want to be able to just create a standard file structure of routings and have a module do it automatically.
花时间调用/定义路由完全没有意义。即使您确实需要自定义控件，也可能只是在某些时候，并且在大多数情况下，您希望能够仅创建路由的标准文件结构并让模块自动执行此操作。
Try Route Magic
尝试路线魔法
As you scale your app, the routing invocations will start to form a giant heap of code that serves no purpose. You want to do just 2 lines of code to handle all the app.userouting invocations with Route Magic like this:
当你扩展你的应用程序时，路由调用将开始形成一大堆毫无用处的代码。你只想做 2 行代码来app.use使用 Route Magic处理所有路由调用，如下所示：
const magic = require('express-routemagic')
magic.use(app, __dirname, '[your route directory]')
For those you want to handle manually, just don't use pass the directory to Magic.
对于那些你想手动处理的，只是不要使用将目录传递给 Magic。
