假设我的示例URL是

http://example.com/one/two

我说我有以下路线

app.get('/one/two', function (req, res) {
    var url = req.url;
}

url的值是/one/two。

如何在Express中获得完整的URL ? 例如,在上面的情况下,我想收到http://example.com/one/two。


当前回答

你可以结合req。协议,要求。主机名和req.originalUrl。请注意要求。主机名,而不是请求。Host或req.get(“Host”),这是可行的,但更难阅读。

const completeUrl = `${req.protocol}://${req.hostname}${req.originalUrl}`;

其他回答

async function (request, response, next) {
  const url = request.rawHeaders[9] + request.originalUrl;
  //or
  const url = request.headers.host + request.originalUrl;
}

The protocol is available as req.protocol. docs here Before express 3.0, the protocol you can assume to be http unless you see that req.get('X-Forwarded-Protocol') is set and has the value https, in which case you know that's your protocol The host comes from req.get('host') as Gopal has indicated Hopefully you don't need a non-standard port in your URLs, but if you did need to know it you'd have it in your application state because it's whatever you passed to app.listen at server startup time. However, in the case of local development on a non-standard port, Chrome seems to include the port in the host header so req.get('host') returns localhost:3000, for example. So at least for the cases of a production site on a standard port and browsing directly to your express app (without reverse proxy), the host header seems to do the right thing regarding the port in the URL. The path comes from req.originalUrl (thanks @pgrassant). Note this DOES include the query string. docs here on req.url and req.originalUrl. Depending on what you intend to do with the URL, originalUrl may or may not be the correct value as compared to req.url.

将这些组合在一起以重建绝对URL。

  var fullUrl = req.protocol + '://' + req.get('host') + req.originalUrl;

下面的代码对我来说就足够了!

const baseUrl = `${request.protocol}://${request.headers.host}`;
// http://127.0.0.1:3333
var full_address = req.protocol + "://" + req.headers.host + req.originalUrl;

or

var full_address = req.protocol + "://" + req.headers.host + req.baseUrl;

我建议使用originalUrl而不是URL:

var url = req.protocol + '://' + req.get('host') + req.originalUrl;

查看originalUrl的描述: http://expressjs.com/api.html#req.originalUrl

在我们的系统中,我们这样做,所以originalUrl对我们很重要:

  foo = express();
  express().use('/foo', foo);
  foo.use(require('/foo/blah_controller'));

Blah_controller是这样的:

  controller = express();
  module.exports = controller;
  controller.get('/bar/:barparam', function(req, res) { /* handler code */ });

我们的url有这样的格式

www.example.com/foo/bar/:barparam

因此,我们需要req。originalUrl在bar控制器的获取处理程序。