使用url.format:

var url = require('url');

这支持所有协议，包括端口号。如果你的originalUrl中没有查询字符串，你可以使用这个更干净的解决方案:

var requrl = url.format({
    protocol: req.protocol,
    host: req.get('host'),
    pathname: req.originalUrl,
});

如果你有一个查询字符串:

var urlobj = url.parse(req.originalUrl);
urlobj.protocol = req.protocol;
urlobj.host = req.get('host');
var requrl = url.format(urlobj);

使用url.format:

var url = require('url');

这支持所有协议，包括端口号。如果你的originalUrl中没有查询字符串，你可以使用这个更干净的解决方案:

var requrl = url.format({
    protocol: req.protocol,
    host: req.get('host'),
    pathname: req.originalUrl,
});

如果你有一个查询字符串:

var urlobj = url.parse(req.originalUrl);
urlobj.protocol = req.protocol;
urlobj.host = req.get('host');
var requrl = url.format(urlobj);

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

or

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

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

我发现它有点PITA，以获得所请求的url。我不敢相信没有更简单的快递方式了。应该是req。requested_url

但我是这样设置的:

var port = req.app.settings.port || cfg.port;
res.locals.requested_url = req.protocol + '://' + req.host  + ( port == 80 || port == 443 ? '' : ':'+port ) + req.path;

你可以从req of express得到完整的url。

function fetchPages(req, res, next) {

    let fullUrl = req.headers.host + req.originalUrl;
    console.log("full url ==> ",fullUrl);
}