Question

我有两种型号，购买和标签购买可以有许多标签（一对多关系）。

什么是最佳（Efficent / Cleanest / etc）返回所有包含一个或多个标签的购买方式？（不必返回实际的标签实体）。

我目前正在购买表中使用一个列来确定它是否已被标记但是想要将其删除（请参阅PurchaseDAO，因为它已被使用）

的 PurchaseController ：

@RequestMapping(value = "purchases/tagged", method = RequestMethod.GET)
@ResponseBody
public final List<Purchase> getTagged()
{
    return RestPreconditions.checkNotNull(purchaseService.getTagged());
}

@RequestMapping(value = "purchases/pending", method = RequestMethod.GET)
@ResponseBody
public final List<Purchase> getPending()
{
    return RestPreconditions.checkNotNull(purchaseService.getPending());
}

的 PurchaseService ：

@Service
public class PurchaseService implements IPurchaseService
{
    @Autowired
    private IPurchaseDAO purchaseDAO;

    public PurchaseService()
    {

    }

    @Transactional
    public List<Purchase> getAll()
    {
        return purchaseDAO.findAll();
    }

    @Transactional
    public List<Purchase> getPending()
    {
        return purchaseDAO.getPending();
    }

    @Transactional
    public List<Purchase> getTagged()
    {
        return purchaseDAO.getTagged();
    }
}

的 PurchaseDAO ：

@Repository
public class PurchaseDAO extends AbstractJpaDAO<Purchase> implements IPurchaseDAO {

    @PersistenceContext
    EntityManager entityManager;

    public PurchaseDAO() {
        setClazz(Purchase.class);
    }

    public  List<Purchase> getPending() {
        return entityManager.createQuery("from Purchase where tagged = 0")
                .getResultList();
    }

    public  List<Purchase> getTagged() {
        return entityManager.createQuery("from Purchase where tagged = 1")
                .getResultList();
    }
}

我的购买和标签DAO都扩展了以下 AbstractJpaDAO ：

public abstract class AbstractJpaDAO<T extends Serializable> implements
        IAbstractJpaDAO<T> {

    private Class<T> clazz;

    @PersistenceContext
    EntityManager entityManager;

    public void setClazz(final Class<T> clazzToSet) {
        this.clazz = clazzToSet;
    }

    public T findOne(final Long id) {
        return entityManager.find(clazz, id);
    }

    public List<T> findAll() {
        return entityManager.createQuery("from " + clazz.getName())
                .getResultList();
    }

    public void save(final T entity) {
        entityManager.persist(entity);
    }
}

由于

JB建议更新

将PurchaseDAO getTagged（）更改为：

public  List<Purchase> getTagged() {
        return entityManager.createQuery("SELECT p FROM Puchase p INNER JOIN p.tags")
                .getResultList();
    }

我收到以下错误：

TRACE [http-bio-8080-exec-3] o.s.w.c.s.AnnotationConfigWebApplicationContext [AbstractApplicationContext.java:322] Publishing event in Root WebApplicationContext: ServletRequestHandledEvent: url=[/PurchaseAPIServer/api/purchases/tagged]; client=[192.168.1.17]; method=[GET]; servlet=[PurchaseAPIServer]; session=[null]; user=[null]; time=[411ms]; status=[failed: org.springframework.dao.InvalidDataAccessApiUsageException: org.hibernate.hql.ast.QuerySyntaxException: Puchase is not mapped [SELECT p FROM Puchase p JOIN p.tags tag]; nested exception is java.lang.IllegalArgumentException: org.hibernate.hql.ast.QuerySyntaxException: Puchase is not mapped [SELECT p FROM Puchase p JOIN p.tags tag]]

Answer 1

select p from Purchase p inner join p.tags

内部联接只会使购买至少有一个标签。

确定父是否有一个或多个使用hibernate的子节点

JB建议更新

1 个答案: