[FIXED] How to get only part of URL from HttpServletRequest?

Issue

From the following URL I need to get (http://localhost:9090/dts) alone.
That is I need to remove (documents/savedoc) (OR)
need to get only - (http://localhost:9090/dts)

http://localhost:9090/dts/documents/savedoc  

Is there any method available in request to get the above?

I tried the following and got the result. But still trying.

System.out.println("URL****************"+request.getRequestURL().toString());  
System.out.println("URI****************"+request.getRequestURI().toString());
System.out.println("ContextPath****************"+request.getContextPath().toString());

URL****************http://localhost:9090/dts/documents/savedoc  
URI****************/dts/documents/savedoc  
ContextPath****************/dts

Can anyone please help me in fixing this?

Solution

AFAIK for this there is no API provided method, need to customization.

String serverName = request.getServerName();
int portNumber = request.getServerPort();
String contextPath = request.getContextPath();

// try this

System.out.println(serverName + ":" +portNumber + contextPath );

Answered By - NPKR
Answer Checked By - Timothy Miller (JavaFixing Admin)