在 Express NodeJS 中调用其他路由中已经定义的路由
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Calling already defined routes in other routes in Express NodeJS
提问by user1148710
I am writing a web app in node.js using Express. I have defined a route as follows:
我正在使用 Express 在 node.js 中编写一个 Web 应用程序。我定义了一条路线如下：
app.get("/firstService/:query", function(req,res){
//trivial example
var html = "<html><body></body></html>";
res.end(html)
});
How do I reuse that route from within express?
如何从 express 中重用该路由？
app.get("/secondService/:query", function(req,res){
var data = app.call("/firstService/"+query);
//do something with the data
res.end(data);
});
I couldn't find anything in the API documentation and would rather not use another library like "request" because that seems kludgey. I am trying to keep my app as modular as possible. Thoughts?
我在 API 文档中找不到任何内容，我宁愿不使用像“请求”这样的另一个库，因为这看起来很笨拙。我试图让我的应用程序尽可能模块化。想法？
Thanks
谢谢
采纳答案by Gates VP
Can you simply break this out into another function, put it in a shared spot and go from there?
你能简单地把它分解成另一个函数，把它放在一个共享的地方然后从那里开始吗？
var queryHandler = require('special_query_handler');
// contains a method called firstService(req, res);
app.get('/firstService/:query', queryHandler.firstService);
// second app
app.get('/secondService/:query', queryHandler.secondService);
Honestly, this whole business of nesting the call back inside of the app.get(...)is not really a great practice. You end up with a giant file containing all of the core code.
老实说，将回调嵌套在里面的整个业务app.get(...)并不是一个很好的做法。您最终会得到一个包含所有核心代码的巨大文件。
What you really want is a file filled with app.get()and app.post()statements with all of the callback handlers living in different, better organized files.
您真正想要的是一个包含所有回调处理程序app.get()和app.post()语句的文件，这些处理程序位于不同的、组织得更好的文件中。
回答by c0deNinja
Similar to what Gates said, but I would keep the function(req, res){}in your routes file. So I would do something like this instead:
类似于盖茨所说的，但我会将其保留function(req, res){}在您的路线文件中。所以我会做这样的事情：
routes.js
路由.js
var myModule = require('myModule');
app.get("/firstService/:query", function(req,res){
var html = myModule.firstService(req.params.query);
res.end(html)
});
app.get("/secondService/:query", function(req,res){
var data = myModule.secondService(req.params.query);
res.end(data);
});
And then in your module have your logic split up like so:
然后在您的模块中将您的逻辑拆分如下：
myModule.js
我的模块.js
var MyModule = function() {
var firstService= function(queryParam) {
var html = "<html><body></body></html>";
return html;
}
var secondService= function(queryParam) {
var data = firstService(queryParam);
// do something with the data
return data;
}
return {
firstService: firstService
,secondService: secondService
}
}();
module.exports = MyModule;
回答by Der Hochstapler
If you have a lot of middleware on your route, you can benefit from spreading:
如果你的路线上有很多中间件，你可以从传播中受益：
const router = express.Router();
const myMiddleware = [
authenticationMiddleware(),
validityCheckMiddleware(),
myActualRequestHandler
];
router.get( "/foo", ...myMiddleware );
router.get( "/v1/foo", ...myMiddleware );
回答by Aminadav Glickshtein
You can use run-middlewaremodule exactly for that
您可以run-middleware完全为此使用模块
app.runMiddleware('/firstService/query',function(responseCode,body,headers){
// Your code here
})
More info:
更多信息：
- Module page in Github& NPM;
- Examples of use run-middleware module
Disclosure: I am the maintainer & first developer of this module.
披露：我是这个模块的维护者和第一个开发者。
回答by Dhiraj
I have used following way: at userpage.js
我使用了以下方式：在 userpage.js
router.createSitemap = function(req, res, callback) { code here callback(value); }
at product.js
在 product.js
var userPageRouter = require('userpages');
userPageRouter.createSitemap(req, res, function () {
//console.log('sitemap');
});
Also can use in same userpage.js router I can use for other routing as well. eg.
也可以在同一个 userpage.js 路由器中使用，我也可以用于其他路由。例如。
router.get('/sitemap', function (req, res, next) {
router.createSitemap(req, res, function () {
res.redirect('/sitemap.xml');
}); });
Hope this will help.
希望这会有所帮助。
