Question

假设我们有这个带JPA注释的类，其属性类型为List。此代码目前正常运行。

@Entity
public class Family {
    ...
    @CollectionOfElements(targetElement=java.lang.String.class)
    @JoinTable(name = "elements_family",
        joinColumns = @JoinColumn(name = "idFamily")
    )
    @Column(name = "element", nullable = false)
    private List<String> elements;
    ...
}

有没有办法查询包含元素“yyy”的系列列表？就是这样：

Query query = getEntityManager().createQuery("select f FROM Family f WHERE element = :element");
query.setParameter("element", "yyy");
return query.getResultList();

Answer 1

请注意，您正在使用Hibernate特定功能CollectionOfElements。这不是严格的JPA。您可以使用HQL特定的elements函数来执行此操作：

select f from Family f WHERE :element in elements(f.elements)

这是一个有效的例子：

import org.hibernate.annotations.CollectionOfElements;

import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.EntityManager;
import javax.persistence.EntityManagerFactory;
import javax.persistence.EntityTransaction;
import javax.persistence.Id;
import javax.persistence.JoinColumn;
import javax.persistence.JoinTable;
import javax.persistence.Persistence;
import javax.persistence.Query;
import java.util.Arrays;
import java.util.List;

    @Entity
    public class Family {
        @Id
        int id;

        String name;

        @CollectionOfElements(targetElement = java.lang.String.class)
        @JoinTable(name = "elements_family",
                joinColumns = @JoinColumn(name = "idFamily")
        )
        @Column(name = "element", nullable = false)
        public List<String> elements;

        public static void main(String[] args) throws Exception {
            EntityManagerFactory emf = 
                Persistence.createEntityManagerFactory("mysql");
            EntityManager em = emf.createEntityManager();
            try {

                Family f1 = new Family();
                f1.id = 1;
                f1.name = "Happy";

                f1.elements = Arrays.asList("foo", "bar", "yyy", "zzz");

                Family f2 = new Family();
                f1.id = 2;
                f2.name = "Disfunctional";
                f2.elements = Arrays.asList("foo", "bar", "yyy", "xxx");

                EntityTransaction tx = em.getTransaction();
                tx.begin();
                em.persist(f1);
                em.persist(f2);
                tx.commit();
                Query query = em.createQuery(
                   "select f from Family f WHERE :element in elements(f.elements)");
                query.setParameter("element", "bar");
                List list = query.getResultList();
                assert list.size() == 2;

            } finally {
                em.close();
                emf.close();
            }
        }
    }

为了完整性，persistence.xml

<persistence-unit name="mysql" transaction-type="RESOURCE_LOCAL">
    <class>jpa.Family</class>
    <properties>
        <property name="hibernate.dialect" value="org.hibernate.dialect.MySQLDialect"/>
        <property name="hibernate.connection.driver_class" 
                          value="com.mysql.jdbc.Driver"/>
        <property name="hibernate.connection.url" value="jdbc:mysql://localhost/test"/>
        <property name="hibernate.connection.username" value="test"/>
        <property name="hibernate.connection.password" value="test"/>
        <property name="hibernate.show_sql" value="true"/>
        <property name="hibernate.format_sql" value="true"/>
        <property name="hibernate.max_fetch_depth" value="3"/>
        <property name="hibernate.bytecode.use_reflection_optimizer" value="true"/>
        <property name="hibernate.archive.autodetection" value="true"/>
        <property name="hibernate.cache.use_second_level_cache" value="true"/>
        <property name="hibernate.generate_statistics" value="true"/>
        <property name="hibernate.hbm2ddl.auto" value="create"/>
    </properties>
</persistence-unit>

Answer 2

JPQL中的MEMBER OF子句：

Query query = getEntityManager().createQuery("select f FROM Family f WHERE :element MEMBER OF f.element"); 
query.setParameter("element", "yyy"); 
return query.getResultList();

Answer 3

List<Family> found = getHibernateTemplate().find(
"from Family as f join f.elements as e where e = ?",
new Object[]{"yyy"});

如何在JPA </string>中查询List <string>类型的属性

3 个答案: