如何从ASP.Net MVC 5应用程序中的浏览器获取URL

Question

I have ASP.Net web application, and I want to fetch the URL present in the browser. Below is my attempt

public static string UrlForGoogleAuth
{
    get
    {
        UrlHelper url = new UrlHelper(HttpContext.Current.Request.RequestContext);

        string host = HttpContext.Current.Request.Url.Host;
        if (host == "localhost")
        {
            host = HttpContext.Current.Request.Url.Authority;
        }
        string result = "https://" + host + url.Action("LoginSettingsSubmit", "Security");
        return result;
    }
}

Scenario 1. Url: https://sample.com - works fine and host variable fetches sample.com

Scenario 2. Url: https://sample.com:2345 - host variable just fetches sample.com instead of sample.com:2345

Work Around (temp fix: hard code 2345 in the result)

 string result = "https://" + host + ":2345" + url.Action("LoginSettingsSubmit", "Security");

But I am looking for a fool proof solution with no hard-coding. Also, If I can achieve the result without changing alot of stuff in already written code, then that would be great.

EDIT:

    public static string UrlForGoogleAuth
    {
        get
        {
            UrlHelper url = new UrlHelper(HttpContext.Current.Request.RequestContext);   
                host = HttpContext.Current.Request.Url.Authority;
            string result = "https://" + host + url.Action("LoginSettingsSubmit", "Security");
            return result;
        }
    }

Answer 1

Uri.Authority will give you the host name and port (if the URL has one).

Gets the Domain Name System (DNS) host name or IP address and the port number for a server.

// www.sample.com:8080
Console.WriteLine(new Uri("https://www.sample.com:8080/controller/action").Authority);

// www.sample.com
Console.WriteLine(new Uri("https://www.sample.com/controller/action").Authority);

So in your case, you would just use the following:

HttpContext.Current.Request.Url.Authority

Answer 2

这将为您提供完整的URL：

var url = Request.RequestContext.HttpContext.Request.RawUrl;