快速路由-根页和404

Question

I am working on a Node app that uses Express. In this app, I have the following:

app.use('/', function(req, res) {
  res.render('index', {});            
}); 

This route works fine for the positive case. However, it fails for the negative case. For example, If I visit " http://www.example.com/404 ", I still see the index page. In reality, I want it to fall through so that the Express Error handler tackles the error.

How do I change my route so that when a person visits the root of my app, they see the home page. yet, if they enter anything else, they'll see the 404 page?

Answer 1

You want to use app.get() (or possibly app.all() ), not app.use() :

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

The reason why app.use('/', ...) matches /404 is explained here :

A route will match any path that follows its path immediately with a “/”. For example: app.use('/apple', ...) will match “/apple”, “/apple/images”, “/apple/images/news”, and so on.

Answer 2

You can define middleware to specifically handle errors.

Express error-handling middleware takes four arguments and should be defined after all your other routes:

app.use('/', function(req, res) {
  res.render('index', {});            
}); 

app.use(function(err, req, res, next) {
  console.error(err.stack);
  res.status(500).send('Something broke!');
});