Question

我有以下条件查询，它会检索Anfrage和Sparte实体中的某些字段以及sparte.i18nKey的已翻译字符串。如果我不使用orderBy，这可以正常工作。

现在我需要按照sparte.i18nKey的翻译字符串排序并使用orderBy，如下所示，结果为QuerySyntaxException: unexpected AST node

所以问题必须是orderBy子句中的子选择！

select distinct new
  my.domain.model.dto.AnfrageDTO(
      anfrage0.id,
      anfrage0.name,
      anfrage0.sparte.id,
      anfrage0.sparte.i18nKey,
      -- retrieve translated string for sparte.i18nKey
      (select rb0.value from at.luxbau.mis2.domain.model.ResourceBundleEntity as rb0
       where (anfrage0.sparte.i18nKey = rb0.key) and (rb0.language = 'de'))
  )
from my.domain.model.impl.Anfrage as anfrage0
  left join anfrage0.sparte as sparte
order by (
  -- sort by translated string for sparte.i18nKey
  select rb1.value
  from my.domain.model.ResourceBundleEntity as rb1
  where (anfrage0.sparte.i18nKey = rb1.key) and (rb1.language = 'de')
) asc

我的Java代码如下所示：

private List<AnfrageDTO> getAnfragen() {
    CriteriaBuilder cb = entityManager.getCriteriaBuilder();
    CriteriaQuery<AnfrageDTO> query = cb.createQuery(AnfrageDTO.class);
    Root<Anfrage> anfrage = query.from(Anfrage.class);  
    anfrage.join(Anfrage_.sparte, JoinType.LEFT);

    query.select(cb.construct(AnfrageDTO.class,
        anfrage.get(Anfrage_.id),
        anfrage.get(Anfrage_.name),
        anfrage.get(Anfrage_.sparte).get(Sparte_.id),
        anfrage.get(Anfrage_.sparte).get(Sparte_.i18nKey),
        // create subquery for translated sparte.i18nKey
        createResourceBundleSubQuery(cb, query, anfrage.get(Anfrage_.sparte).get(Sparte_.i18nKey)).getSelection()));

    TypedQuery<AnfrageDTO> tq = entityManager
            .createQuery(query)
            // use subquery to sort by translated sparte.i18nKey
            .orderBy(cb.asc(createResourceBundleSubQuery(cb, query, anfrage.get(Anfrage_.sparte).get(Sparte_.i18nKey))));

    tq.setMaxResults(10);
    List<AnfrageDTO> anfragen = tq.getResultList();
    return anfragen;
}

public Subquery<String> createResourceBundleSubQuery(CriteriaBuilder cb, CriteriaQuery<?> query, <String> expr) {
    Subquery<String> subquery = query.subquery(String.class);
    Root<ResourceBundleEntity> rb = subquery.from(ResourceBundleEntity.class);
    subquery
        .select(rb.get(ResourceBundleEntity_.value))
        .where(cb.and(
                      cb.equal(expr, rb.get(ResourceBundleEntity_.key)),
                      cb.equal(rb.get(ResourceBundleEntity_.language), "de")));
    return subquery;
}

在orderBy中使用带有subselect的本机SQL查询也可以按预期工作。

select distinct  
  anfrage0_.id,
  anfrage0_.name,
  anfrage0_.sparte_id,
  sparte4_.i18n_key,
  (select rb3.i18n_value from resource_bundle rb3 where rb3.language_code = 'de' and rb3.i18n_key = sparte4_.i18n_key) as sparte_i18n_value
from
  mis2.anfrage anfrage0_
  left outer join mis2.sparte sparte4_ on anfrage0_.sparte_id = sparte4_.id
order by (
  select rb.i18n_value 
  from mis2.resource_bundle rb    
  where sparte4_.i18n_key = rb.i18n_key and rb.language_code = 'de'
) asc

在本机SQL查询中使用别名也可以按预期工作。

select distinct  
  anfrage0_.id,
  anfrage0_.name,
  anfrage0_.sparte_id,
  sparte4_.i18n_key,
  (select rb3.i18n_value from resource_bundle rb3 where rb3.language_code = 'de' and rb3.i18n_key = sparte4_.i18n_key) as sparte_i18n_value
from
  mis2.anfrage anfrage0_
  left outer join mis2.sparte sparte4_ on anfrage0_.sparte_id = sparte4_.id
order by sparte_i18n_value
asc

如果JPA Criteria API支持在orderBy子句中使用别名，那就太棒了！

欢迎任何提示 - 谢谢！

我的环境是WildFly 11和PostgreSQL 9.6。

Answer 1

JPA不支持在order by子句中传递参数，之前已经问过您的问题：Hibernate Named Query Order By parameter

JPA Criteria orderBy：意外的AST节点

1 个答案: