Nodejs Express隐式中间件适用于所有路由吗? [英] Nodejs Express implicit middleware applied to all routes?

问题描述

我想知道Express是否可以让我创建一个路由中间件，而无需我将其显式地放置在app.get()arg列表中吗?

I wanted to know if Express would let me create a route middleware that would be called by default without me explicitly placing it on the app.get() arg list?

//NodeJS newb

// NodeJS newb

var data = { title: 'blah' };

// So I want to include this in every route
function a(){
  return function(req, res){
    req.data = data;
  };
};

app.get('/', function(req, res) {
  res.render('index', { title: req.data.title });
};

我知道我可以执行app.set('data', data)并通过路由中的req.app.settings.data访问它.这可能会满足我上面的简单示例.

I understand I can do app.set('data', data) and access it via req.app.settings.data in the route. Which would probably satisfy my simple example above.

推荐答案

function a(req, res, next){
  req.data = data;
  // Update: based on latest version of express, better use this
  res.locals.data = data;
  next();
};

app.get('/*', a);

请参见快速文档，中间件"部分中的示例.

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