Question

我试图以编程方式创建球衣资源（没有注释）。我有一个方法raiseAlarm，它接受Name和id作为输入参数。我想从JSON输入中获取Name，并且我希望id来自path参数。代码看起来像这样......

public class JerseyExample {
    public static void main(String[] args) {
      JerseyExample deployer = new JerseyExample();
      deployer.init();
    }


public static class BaseResource extends ResourceConfig {
    public BaseResource() {
        init();
    }

    public void init() {
        try {
            Resource.Builder resourceBuilder2 = Resource.builder();
            resourceBuilder2.path("/raiseAlarm/{id}");
            ResourceMethod.Builder method2 = resourceBuilder2.addMethod("POST")
                    .consumes(MediaType.APPLICATION_JSON_TYPE)
                    .produces(MediaType.APPLICATION_JSON_TYPE)
                    .handledBy(this, this.getClass().getMethod("raiseAlarm", Name.class, String.class));

            Resource childResource1 = resourceBuilder2.build();

            Resource.Builder resourceBuilder = Resource.builder();
            resourceBuilder.path("/employee/status");
            resourceBuilder.addChildResource(childResource1);
            Resource rootResource = resourceBuilder.build();
            registerResources(rootResource);
        } catch (Exception e) {
            e.printStackTrace();
        }
    }

    public String raiseAlarm(Name notification,@PathParam("id") String id) {
        System.out.println("INSIDE RAISE ALARM ");
        System.out.println(notification.toString() + " ID: "+id);
        return "Result";
    }

    public void destroy() {

    }

     public static class Name {

        String firstName;
        String lastName;
        String middleName;

        public String getFirstName() {
            return firstName;
        }

        public void setFirstName(String firstName) {
            this.firstName = firstName;
        }

        public String getLastName() {
            return lastName;
        }

        public void setLastName(String lastName) {
            this.lastName = lastName;
        }

        public String getMiddleName() {
            return middleName;
        }

        public void setMiddleName(String middleName) {
            this.middleName = middleName;
        }

        @Override
        public String toString() {
            return firstName + " " + middleName + " " + lastName;
        }
    }
}

public void init() {

    Server server = new Server();

    ServletContextHandler context0 = new ServletContextHandler(ServletContextHandler.SESSIONS);
    ServletHolder serveltHolder1 = new ServletHolder(new ServletContainer(new BaseResource()));

    context0.addServlet(serveltHolder1, "/*");
    context0.setVirtualHosts(new String[]{"@external"});
    ServerConnector connector = new ServerConnector(server);
    connector.setHost("localhost");
    connector.setPort(9069);
    connector.setName("external");

    HandlerCollection collection = new HandlerCollection();
    collection.addHandler(context0);
    server.setHandler(collection);
    server.addConnector(connector);


    try {
        server.start();
        server.join();
    } catch (Exception e) {
        e.printStackTrace();
    }
    }
}

以上代码有效。我想知道一种方式，我可以以编程方式声明路径参数或查询参数，以便我可以将我的方法签名定义为raiseAlarm（名称通知，字符串ID）并避免@PathParam（＆＃34; id＆＃34;）注释

Answer 1

我知道它刚才被问过，但我遇到了同样的问题并找到了答案。请参阅我的问题和自己的答案here以获取完整的代码。

您必须以与@Path注释相同的方式添加带有路径的子资源。之后，您可以通过上下文的getUriInfo()方法获取path参数。

像这样（假设你已经拥有Resource.Builder个对象）：

final Resource.Builder subResourceBuilder = resourceBuilder.addChildResource("{id}");

subResourceBuilder.addMethod("GET")
    .produces(MediaType.APPLICATION_JSON_TYPE)
    .handledBy(new Inflector<ContainerRequestContext, Response>() {

    @Override
    public Response apply(ContainerRequestContext rctx) {
         // Get to the path parameter                    
         MultivaluedMap<String, String> pparams = rctx.getUriInfo().getPathParameters();
         List<String> idValues = pparams.get("id");
         // Create response here
    }
});

以编程方式在Jersey中声明路径参数

1 个答案: