Question

我真的不知道如何用Cassandra的数据创建对象而不破坏我的反应链吗？

我有一些私有方法是整个反应链的一部分：

private Mono<SecurityData> createSecurityData(Security securityOfType) {
    return jobsProgressRepository
        .findByAgentId(securityOfType.getAgentId()) //Flux<JobsProgress>
        .collectList() //Mono<List<JobsProgress>>
        .flatMap(this::getJobsProgressSummary) //Mono<JobsProgressSummary>
        .flatMap(job -> mapToSecurityData(job, securityOfType));
  }

然后我要准备一些对象：

private Mono<SecurityData> mapToSecurityData(JobsProgressSummary job, Security security ) {
    SecurityData securityData = new SecurityData();
    securityData.setAgentId(security.getAgentId());
    securityData.setGroupId(security.getGroupId());
    securityData.setHostname(getHostname(security)); --> here is the problem!!!
    return Mono.just(securityData);
  }

和getHostname方法：

 private String getHostname(Security security) {
    String hostname = "";
    switch(security.getProductType()){
      case VM: hostname = vmRepository
          .findByAgentId(security.getAgentId()).blockFirst().getHostname();
      case HYPER: hostname = hyperRepository
          .findByAgentId(security.getAgentId()).blockFirst().getHostname();
      default: ""
    }
    return hostname;
  }

我的存储库如下：

public interface HostRepository extends ReactiveCassandraRepository<Host, MapId> {
  Flux<Host> findByAgentId(UUID agentId);
}

也许我的方法不对？我当然不能使用

hostRepository
            .findByAgentId(security.getAgentId()).subscribe() // or blockFirst()

因为我不想破坏我的反应链...

我该如何解决我的问题？请毫不犹豫地给出任何甚至很小的提示：）

更新

在这里，我添加了方法getJobsProgressSummary的遗漏正文：

private Mono<JobsProgressSummary> getJobsProgressSummary(List<JobsProgress> jobs) {
    JobsProgressSummary jobsProgressSummary = new JobsProgressSummary();
    jobs.forEach(
        job -> {
          if (job.getStatus().toUpperCase(Locale.ROOT).equals(StatusEnum.RUNNING.name())) {
            jobsProgressSummary.setRunningJobs(jobsProgressSummary.getRunningJobs() + 1);
          } else if (job.getStatus().toUpperCase(Locale.ROOT).equals(StatusEnum.FAILED.name())) {
            jobsProgressSummary.setAmountOfErrors(jobsProgressSummary.getAmountOfErrors() + 1);
          } else if (isScheduledJob(job.getStartTime())) {
            jobsProgressSummary.setScheduledJobs(jobsProgressSummary.getScheduledJobs() + 1);
          }
        });
    Instant lastActivity =
        jobs.stream()
            .map(JobsProgress::getStartTime)
            .map(startTime -> Instant.ofEpochMilli(Long.parseLong(startTime)))
            .max(Instant::compareTo)
            .orElseGet(null);
    jobsProgressSummary.setLastActivity(lastActivity);
    return Mono.just(jobsProgressSummary);
  }

Answer 1

您需要将所有内容链接在一起，当前您的代码就像是命令式和响应式的混合。另外，您永远不需要调用block。

类似下面的事情应该起作用

private Mono<SecurityData> mapToSecurityData(JobsProgressSummary job, Security security ) {
    //Try to get hostname first, then process result
    return getHostname(security)
            //Map it. Probz should use builder or all args constructor to reduce code here
            .map(hostname -> {
                SecurityData securityData = new SecurityData();
                securityData.setAgentId(security.getAgentId());
                securityData.setGroupId(security.getGroupId());
                securityData.setHostname(hostname);
                return securityData;
            });
}

private Mono<String> getHostname(Security security) {
    Mono<String> hostname = Mono.empty();
    switch(security.getProductType()){
        //Also assuming hostname is a field in Security
        //just change Security to class name if not
        case VM: hostname = vmRepository.findByAgentId(security.getAgentId())
            .next()
            .map(Security::getHostname);
        case HYPER: hostname = hyperRepository.findByAgentId(security.getAgentId())
            .next()
            .map(Security::getHostname);
    }
    return hostname;
}

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1 个答案: