你可以使用node.js API来获取url，并将来自express的信息传递给URL.format()，而不是自己将这些东西连接在一起。

例子:

var url = require('url');

function fullUrl(req) {
  return url.format({
    protocol: req.protocol,
    host: req.get('host'),
    pathname: req.originalUrl
  });
}

你需要使用req.headers.host + req.url来构造它。当然，如果你是在不同的端口，你得到的想法;-)

谢谢大家提供这些信息。这是非常烦人的。

把这个添加到你的代码中，你就再也不用考虑它了:

var app = express();

app.all("*", function (req, res, next) {  // runs on ALL requests
    req.fullUrl = req.protocol + '://' + req.get('host') + req.originalUrl
        next()
})

您还可以在那里执行或设置其他操作，例如将日志记录到控制台。

你可以像这样在路由中使用这个函数

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

/**
 * Gets the self full URL from the request
 * 
 * @param {object} req Request
 * @returns {string} URL
 */
const getFullUrl = (req) => `${req.protocol}://${req.headers.host}${req.originalUrl}`;

要求的事情。Protocol会给出HTTP或https， Req.headers.host会给你完整的主机名，比如www.google.com， 要求的事情。originalUrl会给出剩下的路径名(在你的例子中是/one/two)

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;

我的代码是这样的，

params['host_url'] = req.protocol + '：//' + req.headers.host + req.url;