如何在Node/Express中实现HMVC结构 [英] How to implement HMVC structure in Node/Express

问题描述

我的应用程序中有多个"api"端点，前端框架对其AJAX进程使用了​​这些端点.出于组织目的，我想重复使用这些终结点的代码来检索用于服务器端呈现的数据，在某些情况下，我会这样做以与搜索引擎完美配合.因此，理想情况下，我将实现某种HMVC设置以仅访问API端点上的路由.具体来说，我想在发布顶级响应之前获取响应并对其执行其他操作(例如，渲染带有结果的视图).

I have several "api" endpoints in my application that are used by the front-end framework for its AJAX processes. For organizational purposes, I would like to re-use the code for these endpoints to retrieve data for server-side rendering, which I do in some instances to play nicely with search engines. So, ideally I would implement some sort of HMVC setup to simply access a route on the API endpoint. Specifically I would like to get the response and perform additional actions on it before issuing the top-level response (e.g., render a view with results).

例如:

app.get('/post/recent', function(req, res) {
    app.doRequest('/api/posts/', req, function(res2) {
        var data = res2.body;
        res.render('posts/index', data);
    });
});

做到这一点的最佳方法是什么?

What's the best way to do this?

到目前为止，我想出的最好的选择是:

So far the best option I've come up with is:

将端点中的所有逻辑公开为一种方法，该方法将在app.get('...', myFunction)中使用，然后可以在该路径的表达流之外的其他地方调用myFunction. 但是，除非我编写了自己的不依赖于express的中间件实现，否则这将无法为我提供可靠的方式来运行特定于端点的中间件(我也想在HMV​​C请求上运行). .该API端点具有执行if(!hasAccess) res.send(403)之类的中间件，我特别不想在我的主要路线中发生这种情况，因为我想呈现一个漂亮的错误页面，而不仅仅是发送错误代码.

Expose all logic in an endpoint as a method, which would be used in app.get('...', myFunction), and then I could call myFunction elsewhere outside of the express flow for that path. However, this would not give me a reliable way to run middleware specific to the endpoint (which I would also want to run on the HMVC request) unless I wrote my own middleware implementation that did not rely on express. The API endpoint has middleware that does something like if(!hasAccess) res.send(403), which I specifically do NOT want to happen in my main route since I'd want to render a nice error page instead of just sending an error code.

示例:

var getPosts = function(req) {
   var deferred = q.defer()
   doDatabaseQuery(req.query).then(function(response) {
       deferred.resolve(response)
   });
};
app.get('/api/posts', myMiddlewareFunction(), function(req, res) {
   getPosts(req).then(function(response) {
        res.send(response);
   });
);
app.get('/post/recent', function(req, res) {
   // I want to run middleware here, not in root level
   getPosts(req).then(function(response) {
        res.render('post/index', response);
   }, function(err) {
        res.render('error/noaccess');
   });
});

还有更好的主意吗?有没有办法以编程方式访问快速路线并获得结果?

Any better ideas? Is there a way to programmatically access an express route and get the result?

推荐答案

我通过研究Express源代码来解决这个问题.有一种名为handle的方法，可以通过修改后的request和response对象手动调用该方法以获得我想要的效果:

I figured this out by diving into the Express source code. There is a method called handle which I can manually invoke with modified request and response objects to get the effect I want:

app.get '/posts', (req, res) ->
req.url = '/api/posts'
newRes = _.clone(res)
newRes.send = (data, code)->
    if code is 200
        return res.render('posts/index', data)
    else
        return res.render('error')
app.handle(req, newRes)

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