Build URLs with application path part

[Lsync]

Hi guys and sorry for stupid question (can't handle it for an hour yet):

is there an equivalent for
<jstl:url value="${myURL}" />
but for use in controllers? I need an application part too (e. g. URLs like /application/myURL).

Thanks!

[Lsync]

Hmm. I tried this class. Looks very nice! Source thread here.

Maybe more "standard" spring methods exist?

[titiwangsa]

I use implements ServletContextAware

http://static.springsource.org/sprin...textAware.html

ServletContextAware uses "injection by interface", if you implement this interface, you bean would be injected with the servlet context.

Once you get the servlet context (you need at least servlet 2.5), you can call "getContextPath".

http://grepcode.com/file/repo1.maven...xtPath% 28%29

so:
1. add a new member variable

private ServletContext servletContext;

2. implement an interface

public class YourController implements ServletContextAware {
}

3. implement the setter method

4. call

servletContext.getContextPath()

to get the context path.

[Lsync]

titiwangsa, thanks alot for your reply.

With your solution I've found even simpler method (strange, but it was obvious ) - just use request.getContextPath(). Something like this works perfect (I do not know, however, what is the minimum required version of spring for it):

@RequestMapping(value = "/test")
public String testProcess(Model model, HttpServletRequest request) {

	String appPath = request.getContextPath();
	Logger.getLogger(this.getClass()).info("Application path is: " + appPath);

	return "viewName";
}

So, to have full equivalent for <jstl:url value="${myURL}" /> we can use this code:

@RequestMapping(value = "/test")
public String testProcess(Model model, HttpServletRequest request,
		HttpServletResponse response) {

	String myURL = "/nextTest";
	String nextTestFullURL = response.encodeURL(request.getContextPath() + myURL);

	... pass to model, if needed ...

	return "viewName";
}

Best regards
Lsync