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

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

or

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

其他回答

用这个,

var url = req.headers.host + '/' + req.url;

使req.host /点播。当Express后台代理时,主机名有效必须满足两个条件:

App.set('信任代理','环回');在app.js x - forward - host头必须由你自己在webserver中指定。如。apache, nginx

nginx:

server {
    listen myhost:80;
    server_name  myhost;
    location / {
        root /path/to/myapp/public;
        proxy_set_header X-Forwarded-Host $host:$server_port;
        proxy_set_header X-Forwarded-Server $host;
        proxy_set_header X-Forwarded-For $proxy_add_x_forwarded_for;
        proxy_pass http://myapp:8080;
    }
}

apache:

<VirtualHost myhost:80>
    ServerName myhost
    DocumentRoot /path/to/myapp/public
    ProxyPass / http://myapp:8080/
    ProxyPassReverse / http://myapp:8080/
</VirtualHost>
var full_address = req.protocol + "://" + req.headers.host + req.originalUrl;

or

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

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;

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