Java web.xml页面路由

Question

So here is my problem, I am somewhat new to programming, and now I want to make "web.xml" in Java EE project, to rout every url that contains "/profile/" to a profile page, where depending on the id after "/profile/" it'll show current users data. I have the servlet mapped as below...

<servlet-mapping>
    <servlet-name>RoutServlet</servlet-name>
    <url-pattern>profile/*</url-pattern>
</servlet-mapping>

But what I am supposed to do in the servlet so I can get the id, and show current users data? Please give me an advice, or nice resource where I can see how it's done.

Thank you in advance ! :)

Answer 1

That looks fine for the web.xml . In short, if you're using a JAX-RS implementation (eg Jersey ), you should use the @Path annotation with @PathParam for the parameters, so something like:

@Path("/profile")
public class ProfileService {
    @GET
    @Path("/{id}")
    public Profile getProfile(@PathParam("id") String id) {
        //...
    }

See the Oracle guide on RESTful Web services for some more background.

Answer 2

Firstly I would change tha mapping to the following:

<servlet-mapping>
    <servlet-name>RoutServlet</servlet-name>
    <url-pattern>/profile</url-pattern>
</servlet-mapping>

Secondly in RoutServlet you should have a the following code to get the userId

int userId = Integer.parseInt(req.getParameter("userId"));

And finally any URL's that sends a request to RoutServlet should look like this

http://someurl.com/profile?userId=//someId

This way you can let RoutServlet retrieve all the information of the User depending on what Id is in the GET request.

I would let RoutServlet put all the data it retrieves in an ArrayList<Object> profileData , redirect to a JSP and retrieve the data there using Expression Language, but I dont know what you use/prefer to display the data.