如何优化Express.js路由? [英] How to optimize an Express.js route?
问题描述
我正在开发一个具有以下几个页面的保留区域:
/ dashboard
/ dashboard /个人资料
/ dashboard / user
/ dashboard / view
这是一个简单的用户控制面板。目前我有四条路线:
app.all('/ dashboard',function(req,res,next){ / * Code * /});
app.all('/ dashboard / profile',function(req,res,next){/ * Code * /});
app.all('/ dashboard / user',function(req,res,next){/ * Code * /});
app.all('/ dashboard / view',function(req,res,next){/ * Code * /});
我想优化它,因为在上述每个路线中,我必须在开始:
授权(req,function(auth){
if(!auth)return next(errors.fire (403));
/ *路由代码* /
});
此功能检查用户是否已记录,因此我需要在每个保留页面上调用它。 / p>
我会这样做:
app.all('仪表板/ *',功能(req,res,next){
authorized(req,function(auth){
if(!auth)return next(errors.fire(403)) ;
res.render(something,{})
});
});
在res.render调用中的某事
必须是我需要打开的视图(页面)。
我想将其称为 ONE 时间,以删除冗余代码。
这可能是面板的家(如果用户想要/仪表板)或页面(如果用户想要一个页面/仪表板,如/ dashboard / profile)在最后一种情况下,我需要渲染配置文件视图。 p>
(我必须在将视图传递给render()之前进行检查,因为如果有人尝试/ dashboard / blablablabla应该是一个问题。)
谢谢
您可以将该功能作为路由中间件传递到每个路由, http://expressjs.com/guide.html#route-middleware 了解更多信息。这个想法是这样的:
函数mustBeAuthorized(req,res,next){
/ *需要你的代码授权用户* /
}
然后在每个路由中:
app.all('/ dashboard',mustBeAuthorized,function(req,res,next){/ * Code * /});
或者如果您的逻辑依赖于每个路由的某个角色,您可以使路由中间件像这样:
函数mustBeAuthorizedFor(role){
return function(req,res,next){
/ *您的代码需要授权用户使用该ROLE * /
};
}
然后立即致电:
app.all('/ dashboard',mustBeAuthorizedFor('dashboard'),function(req,res,next){/ * Code * /});
I'm developing a reserved area that has the follow few pages:
/dashboard
/dashboard/profile
/dashboard/user
/dashboard/view
that's a simple user control panel. At the moment i have four routes:
app.all('/dashboard', function(req, res, next) { /* Code */ });
app.all('/dashboard/profile', function(req, res, next) { /* Code */ });
app.all('/dashboard/user', function(req, res, next) { /* Code */ });
app.all('/dashboard/view', function(req, res, next) { /* Code */ });
I would like to optimize it because in each of the above routes i have to call this function at the beginning:
authorized(req, function(auth){
if (!auth) return next(errors.fire(403));
/* route code */
});
This function checks if the user is logged, so i need to call it on every reserved page.
I would do something like:
app.all('/dashboard/*', function(req, res, next) {
authorized(req, function(auth){
if (!auth) return next(errors.fire(403));
res.render(something, {})
});
});
the something
inside the res.render call has to be the view (page) I need to open.
I want to call it ONE time, to remove redundant code.
That could be the home of the panel (if the user wants /dashboard) or the page (if the user wants a page inside /dashboard like /dashboard/profile) in the last case i need to render 'profile' view.
(I have to do a check before pass the view to render(), because if someone try /dashboard/blablablabla it should be a problem.)
Thank you
You can pass that function to each route as a route middleware, check http://expressjs.com/guide.html#route-middleware for more info. The idea would be this:
function mustBeAuthorized(req, res, next){
/* Your code needed to authorize a user */
}
And then in each route:
app.all('/dashboard', mustBeAuthorized, function(req, res, next) { /* Code */ });
Or if your logic depends on a certain role for each route, you can make the route middleware like this:
function mustBeAuthorizedFor(role){
return function(req, res, next){
/* Your code needed to authorize a user with that ROLE */
};
}
And then call it right away:
app.all('/dashboard', mustBeAuthorizedFor('dashboard'), function(req, res, next) { /* Code */ });
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