我想从我定义的包含集合的对象中进行查询。对象看起来像这样:
@Entity
public class ValidationLog {
@Embeddable
public static class ValidationLogPK implements Serializable {
private String dataKey;
@Enumerated(EnumType.STRING)
private DataType dataType;
}
@EmbeddedId
private ValidationLogPK id;
@Enumerated(EnumType.STRING)
private ValidationResult validationResult;
@CollectionOfElements(fetch = FetchType.EAGER)
@JoinTable
@Enumerated(EnumType.STRING)
private Set<ValidationRule> validationRules;
}
查询看起来像这样:
"select v.id.dataKey from ValidationLog v " +
" where v.validationResult = :result" +
" and v.id.dataType = :type" +
" and :rule in indices(v.validationRules)"
但是这不起作用。我不确定“索引”功能。问题是我想获取指定类型、结果和规则的所有“dataKeys”。问题是每个“dataKey”可以有很多规则,如您所见......那么我该怎么做呢?
最佳答案
您在寻找成员
吗? JPQL 中是这样的:
SELECT v.id.dataKey
FROM ValidationLog v
WHERE v.validationResult = :result
AND v.id.dataType = :type
AND :rule MEMBER OF v.validationRules
如果您面对HHH-5209 (不确定您是否会这样做,我针对 Hibernate 3.5 报告了此问题),尝试 HQL 变体:
SELECT v.id.dataKey
FROM ValidationLog v
WHERE v.validationResult = :result
AND v.id.dataType = :type
AND :rule IN elements(v.validationRules)
更新:在枚举集合上使用 in elements
时还有另一个问题,即 HHH-5159 。问题不在于查询本身,而在于参数绑定(bind)。使用时:
query.setParameter("rule", ValidationRule.FOO);
Hibernate 绑定(bind)枚举的序列化版本(在我的例子中,查询只返回任何内容)。但是,使用以下内容对我有用:
query.setParameter("rule", ValidationRule.FOO.name());
更新#2:很抱歉,但我认为我无法提供进一步的帮助。当按照 Hibernate 3.4 和 HSQLDB 的建议使用时,我发布的关于枚举集合上的 in element
的内容对我有用。这是测试:
@Test
// http://opensource.atlassian.com/projects/hibernate/browse/HHH-5159
public void testQueryWithInElementOfCollectionOfElementsOfEnums() {
Person person = new Person("Bruce", "Wayne");
Set<SomeEnum> someEnums = new HashSet<SomeEnum>();
someEnums.add(SomeEnum.ONE);
someEnums.add(SomeEnum.TWO);
someEnums.add(SomeEnum.FIVE);
person.setSomeEnums(someEnums);
session.persist(person);
String queryString = "SELECT p FROM Person p WHERE :someEnum in elements(p.someEnums)";
Query query = session.createQuery(queryString);
// query.setParameter("someEnum", SomeEnum.FIVE); // doesn't work, see HHH-5159
query.setParameter("someEnum", SomeEnum.FIVE.name());
List actual = query.list();
assertNotNull(actual);
assertEquals(1, actual.size());
}
和日志:
... 12:40:06.353 [main] DEBUG org.hibernate.SQL - select person0_.id as id11_, person0_.dept as dept11_, person0_.firstName as firstName11_, person0_.gender as gender11_, person0_.lastName as lastName11_ from Person person0_ where ? in ( select someenums1_.element from Person_someEnums someenums1_ where person0_.id=someenums1_.Person_id ) 12:40:06.357 [main] TRACE org.hibernate.type.StringType - binding 'FIVE' to parameter: 1 12:40:06.359 [main] DEBUG org.hibernate.jdbc.AbstractBatcher - about to open ResultSet (open ResultSets: 0, globally: 0) 12:40:06.361 [main] TRACE org.hibernate.type.IntegerType - returning '1' as column: id11_ 12:40:06.363 [main] DEBUG org.hibernate.loader.Loader - result row: EntityKey[com.acme.domain.Person#1]
也许尝试简化查询以缩小问题范围。我不确定它与 in element
部分有关。
引用文献
- JPA 1.0 规范
- 第 4.6.12 节“集合成员表达式”
- Hibernate 核心引用指南
关于java - 使用 JPA/Hibernate 在集合中搜索,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/3940983/