Ben*_*oki 2 java spring spring-rabbit spring-boot
我正在使用 Spring Boot 从服务调用休息控制器方法。当该方法被调用时,我收到错误 java.lang.NullPointerException 广泛的场景是,我的服务从 RabbitMQ 队列接收有效负载并提取有效负载的内容,然后将其传递到控制器以保存到数据库中。队列部分有效(我可以从队列接收消息并提取内容)。数据库部分也可以工作。问题是从服务调用控制器方法。
这是我的休息控制器
@RestController
@RequestMapping("/auth")
public class AuthController implements AuthService {
@Autowired
RabbitTemplate rabbitTemplate;
@Autowired
AuthRepository authRepository;
public AuthModel addAuthenticatable(AuthModel auth){
auth.setCreatedAt(DateTimeUtility.getDateTime());
return authRepository.save(auth);
}
}
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我的服务代码:
public class QueueListener extends AuthController implements MessageListener{
private String identifier;
private JSONArray globalObject;
private int userId;
private String pin;
@Autowired
AuthController authController;
@Override
public void onMessage(Message message) {
String msg = new String(message.getBody());
String output = msg.replaceAll("\\\\", "");
String jsonified = output.substring(1, output.length()-1);
JSONArray obj = new JSONArray(jsonified);
this.globalObject = obj;
this.identifier = obj.getJSONObject(0).getString("identifier");
resolveMessage();
}
public void resolveMessage() {
if(identifier.equalsIgnoreCase("ADD_TO_AUTH")) {
for(int i = 0; i < globalObject.length(); i++){
JSONObject o = globalObject.getJSONObject(i);
this.userId = Integer.parseInt(o.getString("userId"));
this.pin = o.getString("pin");
}
AuthModel authModel = new AuthModel();
authModel.setUserId(userId);
authModel.setPin(pin);
authController.addAuthenticatable(authModel);
}
}
}
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当我调用 AuthController 中的 addAuthenticatable() 方法时,会发生错误。任何帮助将不胜感激。
我希望这不会脱离主题,但通常我们想要实现的是一种洋葱架构。依赖关系应该有一个方向。
您的控制器是您的应用程序的集成点。您希望每个 REST触发某些逻辑的执行。您的控制器不应扩展与业务逻辑有关的类或实现接口。这部分属于另一层。
一切与逻辑有关的东西都属于服务:
@Service
public class AuthService {
@Autowired
private AuthRepository authRepository;
private String attribute;
public boolean isAuthenticated(String username) {
authRepository.doSomething();
//implementation of the logic to check if a user is authenticated.
}
public boolean authenticate(String username, char[] password) {
// implementation of logic to authenticate.
authRepository.authenticate();
}
public AuthModel save(AuthModel model) {
//implementation of saving the model
}
}
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在服务层中提取逻辑,使事物可重用。现在您可以将服务注入controller
@RestController
@RequestMapping("/auth")
public class AuthController {
@Autowired
private AuthService authService;
public AuthModel addAuthenticatable(AuthModel auth){
//process input etc..
return authService.save(auth);
}
}
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或在一个amqListener
@Component
public class QueueListener implements MessageListener {
@Autowired
private AuthService authService;
@Autowired
private SomeOtherService otherService;
@Override
public void onMessage(Message message) {
JSONArray array = processInput();
JSONArray obj = new JSONArray(jsonified);
String identifier = obj.getJSONObject(0).getString("identifier");
// extract the business logic to the service layer. Don't mix layer responsibilities
otherService.doYourThing(obj, identifier);
resolveMessage();
}
private JSONArray processInput(Message message) {
String msg = new String(message.getBody());
String output = msg.replaceAll("\\\\", "");
String jsonified = output.substring(1, output.length()-1);
}
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以及您的配置,以便您可以让 spring 知道在哪里查找带注释的类。
@Configuration
@ComponentScan({"your.service.packages"})
@EntityScan(basePackages = "your.model.package")
@EnableJpaRepositories("your.repository.packages")
@EnableRabbit // probaby
@EnableWebMvc // probably
public class Config {
//you could also define other beans here
@Bean
public SomeBean someBean() {
return new SomeBean();
}
}
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@pvpkiran 给出了您实际问题的答案。但我希望这对你有帮助
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