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Home/ Questions/Q 9233177
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Editorial Team
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Editorial Team
Asked: June 18, 20262026-06-18T06:29:23+00:00 2026-06-18T06:29:23+00:00

I’ve generated a Google Endpoint AppEngine project in Eclipse by using the Generate AppEngine

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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>
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-18T06:29:25+00:00Added an answer on June 18, 2026 at 6:29 am

    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.

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