访问Hello-World Google Cloud Endpoint服务的URL是什么？

Question

I've generated a Google Endpoint AppEngine project in Eclipse by using the Generate AppEngine BackEnd as described in this blog post . What that post does not describe however, and which the official Google Docs describe poorly as well, is which URL I can access that service with locally?

The service generated has one generated endpoint called DeviceInfoEndpoint. The code is shown below as well as the code in web.xml. Which URL should I access listDeviceInfo() with given that I'm hosting on port 8888 locally? I've tried the following:

• http://localhost:8888/_ah/api/deviceinfoendpoint/v1/listDeviceInfo => 404
• http://localhost:8888/_ah/spi/deviceinfoendpoint/v1/listDeviceInfo => 405 GET not supported
• http://localhost:8888/_ah/spi/deviceinfoendpoint/v1/DeviceInfo => 405 GET (...)
• http://localhost:8888/_ah/spi/v1/deviceinfoendpoint/listDeviceInfo = > 405 GET(...)

Exerpt of DeviceInfoEndpoint.java:

@Api(name = "deviceinfoendpoint")
public class DeviceInfoEndpoint {

/**
 * This method lists all the entities inserted in datastore.
 * It uses HTTP GET method.
 *
 * @return List of all entities persisted.
 */
@SuppressWarnings({ "cast", "unchecked" })
public List<DeviceInfo> listDeviceInfo() {
    EntityManager mgr = getEntityManager();
    List<DeviceInfo> result = new ArrayList<DeviceInfo>();
    try {
        Query query = mgr
                .createQuery("select from DeviceInfo as DeviceInfo");
        for (Object obj : (List<Object>) query.getResultList()) {
            result.add(((DeviceInfo) obj));
        }
    } finally {
        mgr.close();
    }
    return result;
}
}

Web.xml:

<?xml version="1.0" encoding="utf-8" standalone="no"?><web-app xmlns="http://java.sun.com/xml/ns/javaee" xmlns:web="http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" version="2.5" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd">

 <servlet>
  <servlet-name>SystemServiceServlet</servlet-name>
  <servlet-class>com.google.api.server.spi.SystemServiceServlet</servlet-class>
  <init-param>
   <param-name>services</param-name>
   <param-value>com.example.dummyandroidapp.DeviceInfoEndpoint</param-value>
  </init-param>
 </servlet>
 <servlet-mapping>
  <servlet-name>SystemServiceServlet</servlet-name>
  <url-pattern>/_ah/spi/*</url-pattern>
 </servlet-mapping>
</web-app>

Answer 1

API request paths should generally conform to the following:

http(s)://{API_HOST}:{PORT}/_ah/api/{API_NAME}/{VERSION}/

If you're interested in fetching/updating/deleting a specific resource, add an ID to the end. In your example, that suggests you should be querying:

http://localhost:8888/_ah/api/deviceinfoendpoint/v1/

(which maps to list when you're making a GET request).

In general, the APIs Explorer available at /_ah/_api/explorer makes it easy to discover and query these URLs.

Answer 2

You can controle the path by use:

  @ApiMethod(path="listDeviceInfo",  httpMethod = HttpMethod.GET)
    public List<DeviceInfo> listDeviceInfo(){
    //... definition

  }

Then you can call that from you client as: http://localhost:8888/_ah/api/deviceinfoendpoint/v1/listDeviceInfo

If you like send parameters then:

@ApiMethod(path="listDeviceInfo",  httpMethod = HttpMethod.GET)
        public List<DeviceInfo> listDeviceInfo(@Named("info") String info){
        //... definition

  }

http://localhost:8888/_ah/api/deviceinfoendpoint/v1/listDeviceInfo?info=holamundo