我开始使用flatiron作为Web应用程序的工具集来设置事物。Flatiron js - 导演 - 如何从表中进行异步路由?
我使用app.plugins.http导演,似乎无法弄清楚如何为静态文件创建“catchall”路线& 404s - 看来.get("<RegEx>")
只与第一个文件夹位置匹配,所以如果<RegEx>
是/.*
,它将匹配/foo
,但不匹配/foo/bar
。
这里是我的代码,一个更好的例子:
在routes.js
:
var routes = {
/* home
* This is the main route, hit by queries to "/"
*/
"/" : {
get: function(){
getStatic("html/index.html",_.bind(function(err,content){
if(err) throw err;
renderContent(this,content);
},this));
}
},
/* static files
* Last rule, if no other routes are hit, it's either a static resource
* or a 404. Check for the file then return 404 if it doesn't exist.
*/
'/(.*)' : {
get : function(){
getStatic(this.req.url,_.bind(function(err,content){
if(!err){
renderContent(this,content);
} else {
this.res.writeHead(404);
// TODO: fancier 404 page (blank currently)
this.res.end();
}
},this))
}
}
}
,并在我的主要的应用程序文件:
/* Define the routes this app will respond to. */
var routes = require('./lib/routes');
/* set up app to use the flatiron http plugin */
app.use(flatiron.plugins.http);
/* loop through routes and add ad-hoc routes for each one */
for(var r in routes){
var route = routes[r];
if(!routes.hasOwnProperty(r)) continue;
for(var method in route){
if(!route.hasOwnProperty(method)) continue;
app.router[method](r,route[method]);
}
}
/* Start the server */
app.listen(8080);
我希望能够保持我的路线在一个单独的模块中,并导入它们 - 我很不清楚这种方法或使用导演和香草http服务器会更好,但我已经尝试了两种方式,没有任何运气。
这里就是我得到:
localhost:8080/
>> (content of index file - this works)
localhost:8080/foo
>> (blank page, 404 header)
localhost:8080/foo/bar
>> (no static file for this - I get a 404 header, but the body is now "undefined" - where is this coming from??)
localhost:8080/css/min.css
>> (this file should exist, but the route is never called. I do however still get a 404 header, and get the "undefined" body)
所以,我假设了“不确定”身体是不确定的路由默认行为。
有没有办法在不为每个深度添加规则的情况下创建一个catchall路线?