How to get only part of URL from HttpServletRequest?

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 );

You say you want to get exactly:

http://localhost:9090/dts

In your case, the above string consist of:

1. scheme: http
2. server host name: localhost
3. server port: 9090
4. context path: dts

(More info about the elements of a request path can be found in the official Oracle Java EE Tutorial: Getting Information from Requests)
##First variant:###

String scheme = request.getScheme();
String serverName = request.getServerName();
int serverPort = request.getServerPort();
String contextPath = request.getContextPath();  // includes leading forward slash

String resultPath = scheme + "://" + serverName + ":" + serverPort + contextPath;
System.out.println("Result path: " + resultPath);

##Second variant:##

String scheme = request.getScheme();
String host = request.getHeader("Host");        // includes server name and server port
String contextPath = request.getContextPath();  // includes leading forward slash

String resultPath = scheme + "://" + host + contextPath;
System.out.println("Result path: " + resultPath);

---

Both variants will give you what you wanted: http://localhost:9090/dts

Of course there are others variants, like others already wrote ...

It's just in your original question you asked about how to get http://localhost:9090/dts, i.e. you want your path to include scheme.

In case you still doesn't need a scheme, the quick way is:

String resultPath = request.getHeader("Host") + request.getContextPath();

And you'll get (in your case): localhost:9090/dts