java - 从 Hibernate persistenceContext 缓存中排除 JPA 实体？

Question

我正在使用 Spring 构建一个 Web 应用程序，并使用 Hibernate 4.3.6 支持的 JPA 来实现持久性。一些背景:作业结果存储在存储库中，并且 ResultsController 对它们进行检查。前端使用长轮询，因此 ResultsController 创建一个 DeferredResult 对象，然后派生一个线程定期检查作业是否完成，以便填充延迟结果并触发响应。

private DeferredResultsResponse getResults(String idString, String runType, boolean returnOnNotDone) {
    String userId = userService.getCurrentUser();

    // Some basic checks; is the ID a valid format, etc. Not relevant,
    // but the "response" variable is set if we find a problem

    final DeferredResultsResponse deferredResult = new DeferredResultsResponse(runId, runType, userId, returnOnNotDone);
    if (response != null) {
        deferredResult.setResult(response);
    } else {
        Thread t = new Thread(() -> completeResult(deferredResult));
        t.run();
    }

    return deferredResult;
}

private void completeResult(final DeferredResultsResponse result) {
    final ResultsIdentifier id = new ResultsIdentifier(result.getJobId(), result.getRunType());
    int attemptCount = 0;
    boolean returnOnUnfinished = result.isReturnOnUnfinished();

    while (!result.hasResult() && attemptCount < MAX_ATTEMPTS) {
        attemptCount++;
 // ------- Problem line: -----------
        Optional<JobStatus> statusMaybe = jobStatusService.get(new ResultsIdentifier(result.getJobId(), result.getRunType()));

        if (!statusMaybe.isPresent()) {
            result.setResult(new ResultsResponse(null, false, null, "Unable to find job status entry."));
            continue;
        }

        JobStatus status = statusMaybe.get();
        // Incomplete job cases: sleep or respond "not done" based on the flag
        if (!status.isComplete() && returnOnUnfinished) {
            result.setResult(new ResultsResponse(false, false, null, null));
            continue;
        } else if (!status.isComplete()) {
            sleep();
            continue;
        }

        // Cases of completion: respond based on success
        // Various business logic of processing results
    }
    if (!result.hasResult()) {
        result.setResult(new ResultsResponse(true, false, null, String.format("Time out after %d checks", MAX_ATTEMPTS)));
    }
}

问题是这样的:问题行上的查询永远不会报告作业状态的更改。经过一番研究，我找到了 Hibernate 的内部结构。在 SessionImpl 中，有一个 StatefulPersistenceContext 类型的字段，它从第一次从数据库中拉出 JobStatus 对象时就保留了该对象的副本。然后，它在同一 session 中的所有后续查询中重用该副本。

现在，我认为可以通过获取当前 session 并调用clear()或refresh(status)来解决这个问题。然而，对我来说，当其他地方都通过 Spring/JPA 存储库进行中介时，必须拉开 JPA 的帷幕并直接使用 Hibernate 的东西是很糟糕的形式。那么，有什么方法可以标记 ORM XML 文件以排除特定类型在 PersistanceContext 中缓存吗？

<小时/>

作为引用，这里是 JobStatus.xml:

<?xml version="1.0" encoding="UTF-8"?>
<entity-mappings xmlns="http://java.sun.com/xml/ns/persistence/orm"
             xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
             xsi:schemaLocation="http://java.sun.com/xml/ns/persistence/orm http://java.sun.com/xml/ns/persistence/orm_2_0.xsd"
             version="2.0">
<entity class="project.model.JobStatus">
<attributes>
  <embedded-id name="jobIdentifier" />

  <basic name="complete" optional="false" />
  <basic name="userId" optional="false" />

  <basic name="successful" optional="true" />
  <basic name="message" optional="true" />

  <basic name="lastUpdateTime" optional="false">
    <temporal>TIMESTAMP</temporal>
  </basic>
</attributes>
</entity>
</entity-mappings>

jobIdentifier 是一个只包含没有子元素的单个元素的元素。

此外，这是带有事务注释的 JobStatusService:

public interface JobStatusService {

/**
 * Retrieve the statuses of all jobs for the current user.
 * @return All jobs' statuses
 */
@Transactional(readOnly = true)
Iterable<JobStatus> getAllByUser();

/**
 * Retrieve the status of a particular job
 * @param identifier the combined job ID and type
 * @return  The persisted job status
 */
@Transactional(readOnly = true, propagation = Propagation.REQUIRES_NEW)
Optional<JobStatus> get(ResultsIdentifier identifier);

/**
 * Save the passed status, subbing in the current user's ID if none is set,
 * and updating the "last updated" time
 * @param status the job status object
 * @return  The persisted status object
 */
@Transactional(readOnly = false)
@Modifying
JobStatus save(JobStatus status);

/**
 * Delete the status of a particular job
 * @param identifier the combined job ID and type
 */
@Transactional(readOnly = false)
@Modifying
void remove(ResultsIdentifier identifier);

/**
 * Remove all stored job statuses for the given user id.
 * @param userId User id
 */
@Transactional(readOnly = false)
@Modifying
void clearByUser(String userId);

Answer 1

the ResultsController creates a DeferredResult object and then spins off a thread

事实上，不，它不是这么做的。你永远不会启动线程。您在同一个线程中执行所有操作，并且只有在 completeResult() 返回后才返回延迟结果。要真正启动一个线程，您必须替换

t.run();

由

t.start();

现在，为了确保问题行始终进入数据库并重新加载作业状态的新值，您应该做的是确保 jobStatusService.get() 在单独交易。使用 Spring，通常通过使用注释方法来完成

@Transactional(propagation = Propagation.REQUIRES_NEW)