简体   繁体   中英

express.Router() get url route with optional parameters

I tried to get optional URL parameter using express.Router(), but it's not working.

If I use app.get, it is working correctly:

app.get('/videos/:category', function(req, res){
    // localhost:9876/videos/music
    debug(req.params); // This is working as expected
});

The only problem is when I try to use like this using express.Router(). I've tried like this:

app.js:

var express = require('express');
var app = express();

var videos = require('./routes/videos');

app.use('/videos/:category', videos);

routes/videos.js:

var express = require('express');
var router = express.Router();

router.get('/:category', function(req, res){
    debug(req.params, req.params.category); // req.params is empty {}
});
module.exports = router;

I've also tried like this:

router.get('/', function(req, res){
        debug(req.params, req.params.category); // req.params is empty {}
});

How to solve this properly? Thanks.

In your both solutions you are getting the wrong route. In the first case it's /videos/:category/:category , in the second case - /videos/:category/ . You need to move parameter to router.js router.get and remove it from app.js app.use :

app.js:

var express = require('express');
var app = express();

var videos = require('./routes/videos');

app.use('/videos', videos);

routes/videos.js:

var express = require('express');
var router = express.Router();

router.get('/:category', function(req, res){
    debug(req.params, req.params.category); // req.params is empty {}
});
module.exports = router;

The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM