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使用JPA，如果给出比连接表更高级的查询，如何连接实体字段？

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我正在使用Hibernate 4.1.3.Final与JPA 2.1和MySQL 5.5.37 . 我有一个具有以下字段的实体：

@Entity
@Table(name = "category",
       uniqueConstraints = { @UniqueConstraint(columnNames = { "NAME" })}
)
public class Category implements Serializable, Comparable<Category> {
    private static final long serialVersionUID = 1L;

    @Id
    @Column(name = "ID")
    @GeneratedValue(generator = "uuid-strategy")
    private String id;

    @NotEmpty
    private Set<Subject> subjects;

    ...
}

没有简单的连接表来链接 subjects 字段，而是有一个稍微复杂的MySQL查询 . 以下是给出特定类别ID的主题的示例：

SELECT DISTINCT e.subject_id 
FROM category c, resource_category rc, product_resource pr, 
     sb_product p, product_book pe, book e 
WHERE c.id = rc.category_id 
  AND rc.resource_id = pr.resource_id 
  AND pr.product_id = p.id 
  AND p.id = pe.product_id 
  AND pe.ebook_id = e.id 
  AND c.id = ‘ABCEEFGH‘;

在加载类别时，使用下面的查询连接上述字段的最简单方法是什么？

这个问题涉及处理Java以实现这一点，因此构建视图或做其他类型的MySQL疯狂不是一种选择，至少作为这个问题的答案 .

Edit:

根据建议添加了表示法（用'= id'替换'='ABCDEFG“'）但是当我查询绑定到Category实体的项目时，Hibernate会生成这个无效的SQL . 这是SQL Hibernate吐出来的

SELECT categories0_.resource_id AS RESOURCE1_75_0_,
       categories0_.category_id AS CATEGORY2_76_0_,
       category1_.id            AS ID1_29_1_,
       category1_.NAME          AS name2_29_1_,SELECT DISTINCT e.subject_id
FROM            category c,
                resource_category rc,
                product_resource pr,
                product p,
                product_ebook pe,
                book e
WHERE           c.id = rc.category_id
AND             rc.resource_id = pr.resource_id
AND             pr.product_id = p.id
AND             p.id = pe.product_id
AND             pe.ebook_id = e.id
AND             c.id = category1_.id as formula1_1_,
                subject2_.id         AS id1_102_2_,
                subject2_.NAME       AS name2_102_2_
FROM            resource_category categories0_
INNER JOIN      category category1_
ON              categories0_.category_id=category1_.id
LEFT OUTER JOIN subject subject2_
ONSELECT DISTINCT e.subject_id
FROM            category c,
                resource_category rc,
                product_resource pr,
                product p,
                product_ebook pe,
                book e
WHERE           c.id = rc.category_id
AND             rc.resource_id = pr.resource_id
AND             pr.product_id = p.id
AND             p.id = pe.product_id
AND             pe.ebook_id = e.id
AND             c.id = category1_.id=subject2_.id
where           categories0_.resource_id=?

注意结尾处的“左外连接主题subject2_在SELECT DISTINCT e.subject_id上”和“AND c.id = category1_.id = subject2_.id” .

Edit 2 ：

以下是上述查询中涉及的实体

@Entity
@Table(name="resource")
public class Resource implements Serializable, Comparable<Resource>
{
    private static final long serialVersionUID = 1L;

    @Id
    @GeneratedValue(generator = "uuid-strategy")
    private String id;

    …    
    @Column(name = "FILE_NAME")
    private String fileName;

    @Column(name = "URI")
    private String uri;

    …

    @ManyToMany(fetch = FetchType.EAGER)
    @JoinTable(name = "resource_category", joinColumns = { @JoinColumn(name = "RESOURCE_ID") }, inverseJoinColumns = { @JoinColumn(name = "CATEGORY_ID") })
    private Set<Category> categories;

这是查询本身......

CriteriaBuilder builder = m_entityManager.getCriteriaBuilder();
CriteriaQuery<T> criteria = builder.createQuery(Resource.class);
Root<T> rootCriteria = criteria.from(Resource.class);
criteria.select(rootCriteria).where(builder.equal(rootCriteria.get(“uri”),uri));
Resource ret = null;
try {
        final TypedQuery<T> typedQuery = m_entityManager.createQuery(criteria);
        ret = typedQuery.getSingleResult();
} catch (NoResultException e) {
        LOG.warn(e.getMessage());
}
return ret;
java mysql hibernate jpa hibernate-mapping

2 回答

  • 4

    你需要使用Hibernate特定的JoinColumnOrFormula：

    public class Category implements Serializable, Comparable<Category> {
    private static final long serialVersionUID = 1L;

    @Id
    @Column(name = "ID")
    @GeneratedValue(generator = "uuid-strategy")
    private String id;

    @NotEmpty
    @ManyToOne
    @JoinColumnsOrFormulas({
        @JoinColumnOrFormula(
            formula = @JoinFormula(
                value = 
                    "SELECT DISTINCT e.subject_id " +
                    "FROM category c, resource_category rc, product_resource pr, " +
                    "     sb_product p, product_book pe, book e " +
                    "WHERE c.id = rc.category_id " +
                    "  AND rc.resource_id = pr.resource_id " +
                    "  AND pr.product_id = p.id " +
                    "  AND p.id = pe.product_id " +
                    "  AND pe.ebook_id = e.id " +
                    "  AND c.id = ‘ABCEEFGH‘", 
                referencedColumnName="id"
            )
        )
    })
    private Set<Subject> subjects;

    ...
}

    编辑

    您可以在存储过程中包含此查询：

    CREATE FUNCTION join_book(text) RETURNS text
    AS  'SELECT DISTINCT e.subject_id ' +
        'FROM category c, resource_category rc, product_resource pr, ' +
        '     sb_product p, product_book pe, book e ' +
        'WHERE c.id = rc.category_id ' +
        '  AND rc.resource_id = pr.resource_id ' +
        '  AND pr.product_id = p.id ' +
        '  AND p.id = pe.product_id ' +
        '  AND pe.ebook_id = e.id ' +
        '  AND c.id = $1;'  
    LANGUAGE SQL
    IMMUTABLE
    RETURNS NULL ON NULL INPUT;

    然后您的映射变为：

    public class Category implements Serializable, Comparable<Category> {
    private static final long serialVersionUID = 1L;

    @Id
    @Column(name = "ID")
    @GeneratedValue(generator = "uuid-strategy")
    private String id;

    @NotEmpty
    @ManyToOne
    @JoinColumnsOrFormulas({
        @JoinColumnOrFormula(
            formula = @JoinFormula(
                value = 
                    "join_book(id)", 
                referencedColumnName="id"
            )
        )
    })
    private Set<Subject> subjects;

    ...
}
    回复于
  • 1

    SessionFactory范围的hibernate拦截器实现可能会有所帮助 . 我没有一个有效的例子 .

    回复于

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