如何创建/自定义Spring启动执行器为我的Camunda应用程序公开应用程序健康检查和诊断？

Question

我被要求为我的Camunda Engine/BPMN工具创建Spring靴执行器。因此，我们可以通过SpringActuator公开各种应用程序的健康检查和诊断。

这应该扩展到公开Camunda的运行状况、配置和指标。你知道怎么做吗？

Answer 1

我已经能够定制现有的healthcheck执行器。为希望在这些项目中进行自定义的人共享下面的代码。

@Component
public class X extends AbstractHealthIndicator {
    @Autowired
    public JobExecutor jobExecutor;

    @Override
    protected void doHealthCheck(Builder builder) throws Exception {
        boolean active = jobExecutor.isActive();
        if (active) {
            builder = builder.up();
        } else {
            builder = builder.down();
        }
        builder.withDetail("jobExecutor", healthDetails(jobExecutor));
    }

    private Details healthDetails(JobExecutor jobExecutor) {
        final DetailsBuilder builder = Details.builder().name(jobExecutor.getName())
                .lockOwner(jobExecutor.getLockOwner()).lockTimeInMillis(jobExecutor.getLockTimeInMillis())
                .maxJobsPerAcquisition(jobExecutor.getMaxJobsPerAcquisition())
                .waitTimeInMillis(jobExecutor.getWaitTimeInMillis());

        for (ProcessEngineImpl processEngineImpl : jobExecutor.getProcessEngines()) {
            builder.processEngineName(processEngineImpl.getName());
        }

        return builder.build();
    }

}

Answer 2

假设您使用的是camunda spring boot启动器：启动器中包含ProcessEngine和JobExecutor的两个默认运行状况指示器，请查看：

https://github.com/camunda/camunda-bpm-spring-boot-starter/tree/master/starter/src/main/java/org/camunda/bpm/spring/boot/starter/actuator

 @Override
  protected void doHealthCheck(Builder builder) throws Exception {
    boolean active = jobExecutor.isActive();
    if (active) {
      builder = builder.up();
    } else {
      builder = builder.down();
    }
    builder.withDetail("jobExecutor", Details.from(jobExecutor));
  }

这将让您了解如何实现springsAbstractHealthIndicator，访问camunda组件并使用它们进行健康检查。