Strange express.use() behavior in NodeJS

Question

app.use(function(req,res,next){ console.log('middleware executed'); next(); });
app.get('/1',function(req,res){ console.log('/1'); res.end(); });
app.get('/2',function(req,res){ console.log('/2'); res.end(); });
...

It works. Middleware executed on both /1 and /2 requests.

Now I want middleware to not be executed on request page /1, but on all pages below:

app.get('/1',function(req,res){ console.log('/1'); res.end(); });
app.use(function(req,res,next){ console.log('middleware executed'); next(); });
app.get('/2',function(req,res){ console.log('/2'); res.end(); });
...

In this case middleware won't execute. Even on request page /2. What's wrong? How to solve this?

robertklep · Accepted Answer · 2014-02-18 14:18:30Z

2

The reason is that once you declare a route, Express will (under the hood) insert the router middleware into the middleware chain. This means that any incoming requests will get handled by the router and will (probably) not get passed to any middleware declared after the router/route.

@jgitter's suggestion of including the middleware in the route declaration will work just fine.

Another option is to use app.all instead of app.use:

app.get('/1',function(req,res){ console.log('/1'); res.end(); });
app.all('*', function(req,res,next){ console.log('middleware executed'); next(); });
app.get('/2',function(req,res){ console.log('/2'); res.end(); });

edited Feb 18, 2014 at 14:18

answered Feb 18, 2014 at 14:02

robertklep

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1 Comment

E_net4 Over a year ago

You probably mean app.use instead of all.use. ;)

jgitter · Accepted Answer · 2014-02-18 13:14:33Z

1

That looks correct to me, I'm not sure why it isn't working offhand. As a workaround, you can do this instead:

var middleware = function (req, res, next) { 
    // do things
    next();
};
app.get('/1', middleware, function(req, res) { /* handle request */ });
app.get('/2', function(req, res) { /* handle request */ });

You can also add a mount path as the optional first parameter to the app.use() method.

answered Feb 18, 2014 at 13:14

jgitter

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