使用ExpressJS重写URL
使用ExpressJS重写URL
提问于 2013-05-16 13:11:35
我想重写我在ExpressJS网站上的URL。我用过这个插件，https://github.com/joehewitt/express-rewrite，但它不工作...
然而，我可能犯了一个错误...
我的app.js文件：
var express = require('express')
, index = require('./routes/index.js')
, admin = require('./routes/admin.js')
, contact = require('./routes/contact.js')
, posts = require('./routes/posts.js')
, http = require('http')
, path = require('path')
, hash = require('./auth').hash
, db = require('./models')
, favicons = require('connect-favicons')
, rewriter = require('express-rewrite');
var app = express();
app.configure(function () {
app.set('port', process.env.PORT || 3000);
app.set('views', __dirname + '/views');
app.set('view engine', 'jade');
app.use(express.favicon(__dirname + '/public/images/FAVICON.ico'));
app.use(favicons(__dirname + '/public/images/apple-touch-icon.png'));
app.use(express.logger('dev'));
app.use(express.bodyParser());
app.use(express.static(path.join(__dirname, 'public')));
app.use(express.cookieParser());
app.use(express.cookieSession({
secret: 'SECRET',
cookie: { access: false }
})
);
app.use(rewriter);
app.use(app.router);
app.use(function(req, res, next){
res.render('404.jade', {
title: "404 - Page Not Found",
showFullNav: false,
status: 404,
url: req.url
});
});
});
app.configure('development', function () {
app.use(express.errorHandler());
});
app.get('/', index.index);
app.get('/toto', rewriter.rewrite('/heytoto'));
db.sequelize.sync().complete(function(err) {
if (err) {
throw err
} else {
http.createServer(app).listen(app.get('port'), function(){
console.log('Express server listening on port ' + app.get('port'))
})
}
});我的错误消息：
Express
500 TypeError: Object function app(req, res){ app.handle(req, res); } has no method 'match'
at Object.rewriter [as handle] (/Users/anthonycluse/Sites/Anthony-Cluse-Express/node_modules/express-rewrite/rewrite.js:3:26)
at next (/Users/anthonycluse/Sites/Anthony-Cluse-Express/node_modules/express/node_modules/connect/lib/proto.js:199:15)
at Object.cookieSession [as handle] (/Users/anthonycluse/Sites/Anthony-Cluse-Express/node_modules/express/node_modules/connect/lib/middleware/cookieSession.js:113:5)
at next (/Users/anthonycluse/Sites/Anthony-Cluse-Express/node_modules/express/node_modules/connect/lib/proto.js:199:15)
at Object.cookieParser [as handle] (/Users/anthonycluse/Sites/Anthony-Cluse-Express/node_modules/express/node_modules/connect/lib/middleware/cookieParser.js:60:5)
at next (/Users/anthonycluse/Sites/Anthony-Cluse-Express/node_modules/express/node_modules/connect/lib/proto.js:199:15)
at resume (/Users/anthonycluse/Sites/Anthony-Cluse-Express/node_modules/express/node_modules/connect/lib/middleware/static.js:60:7)
at SendStream.error (/Users/anthonycluse/Sites/Anthony-Cluse-Express/node_modules/express/node_modules/connect/lib/middleware/static.js:73:37)
at SendStream.EventEmitter.emit (events.js:126:20)
at SendStream.error (/Users/anthonycluse/Sites/Anthony-Cluse-Express/node_modules/express/node_modules/send/lib/send.js:147:51)回答 6
Stack Overflow用户
回答已采纳
发布于 2013-06-22 15:18:42
所以我也遇到了同样的问题。我写了一个在浏览器上使用历史API的应用程序，我想把所有的非静态URL重写回index.html。所以对于静态文件，我这样做了：
app.configure(function() {
app.use('/', express.static(__dirname + '/'));
});但是对于历史API生成的路径，我这样做了：
app.get('*', function(request, response, next) {
response.sendfile(__dirname + '/index.html');
});这意味着不会重写或重定向任何不是/中的文件或目录的请求(例如历史API生成的URL )，而是为index.html文件提供服务，然后执行所有JS路由魔术。
希望这接近你正在寻找的东西？
Stack Overflow用户
发布于 2014-01-08 03:18:02
您可以在到达要使用的处理程序之前重写URL。
app.use(function(req, res, next) {
if (req.url === '/toto') {
req.url = '/heytoto';
}
next();
});
app.get('/heytoto', ...);我使用了一个similar method来用正则表达式重写网址。
Stack Overflow用户
发布于 2014-04-10 00:14:58
没有response.sendfile(..)的解决方案是使用在app.use(express.static(..))之前插入的重写中间件，如下所示：
// forward all requests to /s/* to /index.html
app.use(function(req, res, next) {
if (/\/s\/[^\/]+/.test(req.url)) {
req.url = '/index.html';
}
next();
});
// insert express.static(...)这样，expressjs就可以正确地识别重写。然后，static中间件将负责为文件提供服务。
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原文链接：
https://stackoverflow.com/questions/16579404
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