如何使用Java配置创建Spring RESTFul服务？

Question

我试图使用Java配置创建一个Spring4 RESTFul服务，并将其部署到Tomcat。但我无法到达终点。我错过了什么？这就是我所拥有的。

public class Greeting {
    private BigInteger id;
    private String text;

    //setters and getters
}

@Controller
public class GreetingController {

  private static BigInteger nextId;
  private static Map<BigInteger, Greeting> greetingMap;

  //have some code to store Greetings in greetingMap

  @RequestMapping(value = "/api/greetings", method = RequestMethod.GET, produces = MediaType.APPLICATION_JSON_VALUE)
  public ResponseEntity<Collection<Greeting>> getGreetings() {
      Collection<Greeting> greetings = greetingMap.values();
      return new ResponseEntity<Collection<Greeting>>(greetings, HttpStatus.OK);
  }

}

@Configuration
@ComponentScan(basePackages = "org.example")
  public class GreetingConfiguration {

}

<web-app id="WebApp_ID" version="2.4"
xmlns="http://java.sun.com/xml/ns/j2ee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee 
http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd">

  <display-name>Spring Application</display-name>

  <servlet>
      <servlet-name>dispatcher</servlet-name>
      <servlet-class>org.springframework.web.servlet.DispatcherServlet</servlet-class>
      <init-param>
          <param-name>contextClass</param-name>
          <param-value>org.springframework.web.context.support.AnnotationConfigWebApplicationContext</param-value>
      </init-param>
      <init-param>
          <param-name>contextConfigLocation</param-name>
          <param-value>org.example.ws.configuration.GreetingConfiguration</param-value>
      </init-param>
      <init-param>
          <param-name>dispatchOptionsRequest</param-name>
          <param-value>true</param-value>
      </init-param>
      <load-on-startup>1</load-on-startup>
  </servlet>

  <context-param>
      <param-name>contextConfigLocation</param-name>
      <param-value>org.example.ws.configuration.GreetingConfiguration</param-value>
  </context-param>

  <servlet-mapping>
      <servlet-name>dispatcher</servlet-name>
      <url-pattern>/</url-pattern>
  </servlet-mapping>

  <listener>
      <listener-class>org.springframework.web.context.ContextLoaderListener</listener-class>
  </listener>

我被派往Tomcat。Tomcat开始很好。我试着点击它http://localhost:8080/api/greetings，它给了我404。我错过了什么？

谢了！

Answer 1

您正在结合XML和Java配置。因此，您需要告诉Spring通过以下两种方法之一来获取控制器注释：

1. 配置类中的@enableWebMVC
2. 或，用XML表示