node.js - 递归 Expressjs 路由

Question

在不求助于正则表达式的情况下，expressjs 可以递归调用路由，即 url 示例:

/f:forum/s:section/t:thread/p:post  
/f:forum/s:section/s:section/t:thread/p:post  
/f:forum/s:section/s:section/s:section/t:thread/p:post  
...  

因此在技术上允许论坛中无限数量的“部分/子部分”。

我尝试这样做:

应用程序.js:

var express = require('express');
app = express();
app.route('/').get(function(req, res, next){
    return res.send('hello');
});
app.use('/f:forum', require('./section'));
server = app.listen(process.env.http || process.env.PORT);
module.exports = app;

section.js:

var router = require('express').Router();
router = router;
router.route('/s:section').get(function(req, res, next){
  return res.send(req.params);
});
router.use('/s:section', require('./thread'));
module.exports = router;

线程.js:

var router = require('express').Router();
router.use('/s:section', require('./section'));
router.route('/t:thread/p-:post').get(function(req, res, next){
  return res.send(req.params);
});
router.route('/t:thread').get(function(req, res, next){
  return res.send(req.params);
});
module.exports = router;

但有趣的是，它告诉我在 thread.js 中 require('./section') = {}
但在 app.js 中它是正确的...有什么建议吗？

Answer 1

您可以像 router.route('/:path*') 那样执行通配符路由，然后让处理程序从该点向下解析。

例如，像这样的东西:

router.route('/forum/:path*', function(req,res){
   var requestPath = req.path; // will present the whole path to you for parsing
   // do whatever db lookup logic you normally would do now that you have the pieces you wanted
   res.render('forum', data);
};