递归Expressjs路由

Question

没有求助于正则表达式反正是有与expressjs递归调用路由即URL例子：

/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 
... 

因此技术上允许在论坛“分段/小分段”的无限量。

我试图做的事：

app.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; 

thread.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 require('./section') = {}
尚未在app.js中，这是正确的...任何建议？

Answer 1

您可以执行通配符路由，如router.route('/:path*')，然后从那一点开始解析处理程序。

例如，像：

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); 
};