I’m using Hibernate 4.1.3.Final with JPA 2.1 and MySQL 5.5.37. I have an entity with the following field:
@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;
...
}
There is no simple join table to link up the subjects
field, instead, there is a slightly more complex MySQL query. Here is an example of figuring out the subjects given a particular category id:
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‘;
What is the simplest way to wire up the above field using the query below when loading categories?
This question concerns dealing with Java to accomplish this so building a view or doing some other type of MySQL madness is not an option, at least as an answer for this question.
Edit:
Added the notation per the suggestion (replacing '="ABCDEFG"' with '=id') but Hibernate is generating this invalid SQL when I do queries for items tied to the Category entity. Here is the SQL Hibernate spits out
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=?
Notice the "left outer join subject subject2_ on SELECT DISTINCT e.subject_id" and "AND c.id = category1_.id=subject2_.id" towards the end.
Edit 2:
Here is the entity involved in the above query
@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;
and here is the query itself …
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;
It is not necessary to have @JoinColumn annotation. You can always override it. If you won't provide it in your code then Hibernate will automatically generate one for you i.e. default name for your column. Could you please be more specific?
Create an entity class Student. java under com. javatpoint. mapping package that contains student id (s_id), student name (s_name) with @OneToMany annotation that contains Library class object of List type.
Association mappings are one of the key features of JPA and Hibernate. They model the relationship between two database tables as attributes in your domain model. That allows you to easily navigate the associations in your domain model and JPQL or Criteria queries.
You need to use Hibernate specific 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;
...
}
Or, you can include this query in a stored procedure:
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;
And, then your mapping becomes:
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;
...
}
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