类实现EnvironmentPostProcessor时如何创建bean

Question

我想为此覆盖/actuator/health，我编写了一个自定义类（CustomHealth.java）并覆盖

getHealth()

但是我无法创建Bean，所以当我调用/actuator/health时，它必须调用我的自定义类实施。

自定义环境后处理器

Class CustomEnvironemtPostProcessor implements EnvironmentPostProcessor {

    public void postProcessEnvironment(ConfigurableEnvironment environment, SpringApplication application) {
        application.addInitializers(new CustomApplicationContextInitializer());
       .....
   
    }
}

自定义应用程序上下文初始化器

class CustomApplicationContextInitializer implements  ApplicationContextInitializer {

@Override
    public void initialize(ConfigurableApplicationContext applicationContext) {       
        ConfigurableListableBeanFactory beanFactory
                = applicationContext.getBeanFactory();
        beanFactory.registerSingleton("customHealth", CustomHealth.class);
    }
}

定制健康

公共类 CustomHealth 实现 HealthIndicator {

public Health getHealth(boolean includeDetails) {
    return this.health();
}

@Override
public Health health() {
    double chance = ThreadLocalRandom.current().nextDouble();
    Health.Builder status = Health.up();
    if (chance > 0.9) {
        status = Health.down();
    }
    return status.build();
}

}

Answer 1

通常您不需要为健康指标创建如此复杂的设置：

只需实现

HealthIndicator

接口并将其注册为Spring bean即可：

@Component
public class RandomHealthIndicator implements HealthIndicator {

    @Override
    public Health health() {
         double chance = ThreadLocalRandom.current().nextDouble();
         Health.Builder status = Health.up();
         if (chance > 0.9) {
            status = Health.down();
         }
         return status.build();
    }
}

代替

@Component

，您可以使用配置方法：

// no @Component annotation here
public class RandomHealthIndicator implements HealthIndicator {
...
}

@Configuration
public class MyHealthIndicatorExtensionsConfiguration {

   @Bean
   public RandomHealthIndicator randomHealthIndicator() {
      return new RandomHealthIndicator();
   }
}

有关完整示例和说明，请参阅此链接。