How to get only part of URL from HttpServletReques

问题:

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?

回答1:

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 );

回答2:

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

回答3:

Just remove URI from URL and then append context path to it. No need to fiddle with loose schemes and ports which is only more tedious when you're dealing with default port 80 which don't need to appear in URL at all.

StringBuffer url = request.getRequestURL();
String uri = request.getRequestURI();
String ctx = request.getContextPath();
String base = url.substring(0, url.length() - uri.length() + ctx.length());
// ...

See also:

• Browser can't access/find relative resources like CSS, images and links when calling a Servlet which forwards to a JSP (for the JSP/JSTL variant of composing the base URL)

回答4:

In my understanding, you need the domain part and Context path only. Based on this understanding, You can use this method to get the required string.

String domain = request.getRequestURL().toString();
String cpath = request.getContextPath().toString();

String tString = domain.subString(0, domain.indexOf(cpath));

tString = tString + cpath;