How to get only part of URL from HttpServletRequest?

Question

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 on

Answer 1

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);

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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