JPA Criteria API - Accessing fields of One-to-Many collection association - jpa

Question

How could I write the following JPQL query using Criteria API?
" select a from A a left join fetch a.bs b where b.i like 'X%' "
A-to-B is a One-To-Many relationship where A and B is like following:
#Entity
public class A {
#OneToMany(cascade={CascadeType.PERSIST})
private Set<B> bs = new HashSet<B>();
//...
}
#Entity
public class B {
private String i;
//...
}
I tried the foolowing:
CriteriaBuilder cb = em.getCriteriaBuilder();
CriteriaQuery<A> cq = cb.createQuery( A.class );
Root<A> aRoot = cq.from( A.class );
Fetch<A, B> bs = aRoot.fetch(A_.bs, JoinType.LEFT);
cq.where(cb.like(what_do_i_do_here, "X%"));
cq.select(aRoot);
I need to get all those As along with its associated Bs where the i value of the associated Bs start with an "X".
EDIT:
If I try the method given at How to properly express JPQL "join fetch" with "where" clause as JPA 2 CriteriaQuery? I incorrectly get 2 As in result while I should get just 1 A whose associated B_.i has the value that starts with "X".
I got the issue resolved using the following code:
CriteriaBuilder cb = em.getCriteriaBuilder();
CriteriaQuery<A> cq = builder.createQuery( A.class );
Root<A> root = cq.from( A.class );
Join<A, B> bs = (Join<A, B>) root.fetch(Guide_.students, JoinType.LEFT);
cq.where(builder.like(bs.get(B_.i), builder.parameter(String.class, "i")));
cq.select(root);
TypedQuery<A> query = em.createQuery(criteria).setParameter("i", "X%");
List<A> as= query.getResultList();
for (A a: as) {
System.out.println(a);
}

Answer 1

You can use Join<ParentTable,ChildTable> to get and filter child entities. Try below code
CriteriaBuilder cb = em.getCriteriaBuilder();
CriteriaQuery<A> cq = cb.createQuery( A.class );
Root<A> aRoot = cq.from( A.class );
Join<A, B> bs = root.join(A_.B);
cq.where(cb.like(bs.get(B_.i), "X%"));
cq.select(aRoot);
Note - I didn't get a chance to test this code. You can refer to working code which I implemented for similar problem here https://github.com/bsridharpatnaik/SpecificationTest
Edit - Providing Detailed Answer
I have made new push to the repo. Please take latest pull.
I created exact same example that you mentioned.
Entity classes
public class A
{
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
Long aid;
#OneToMany(cascade={CascadeType.PERSIST})
private Set<B> bs = new HashSet<B>();
// get & set
}
public class B
{
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
Long bid;
String i;
//get & set
}
Now, as per your query, you want to filter all records from class A where corresponding B.i starts with x.
Below is service class. to make it easy to understand, I wrote criteroa query logic in separate class
#Service
public class Service1
{
#Autowired
aRepo repo;
#Autowired
ModelSpecification modelSpecification;
#Autowired
EntityManager entityManager;
public List<?> getResults() throws ParseException
{
//ModelSpecification modelSpecification = new ModelSpecification();
CriteriaQuery<A> query = modelSpecification.getSpecQuery();
TypedQuery<A> typedQuery = entityManager.createQuery(query);
List<A> resultList = typedQuery.getResultList();
return resultList;
}
}
Below is the criteria query logic
#Component
public class ModelSpecification
{
#Autowired
EntityManager entityManager;
public CriteriaQuery<A> getSpecQuery() throws ParseException
{
CriteriaBuilder builder = entityManager.getCriteriaBuilder();
CriteriaQuery<A> query = builder.createQuery(A.class);
Root<A> root = query.from(A.class);
Join<A,B> ab = root.join(A_.BS);
query.where(builder.like(ab.get(B_.I),"x"+"%"));
return query;
}
}
I am filtering all A records where B_.I starts with x.
Output
Class A has below entries for aid - 1,2
Class B has below entries for
join table a_bs has below entries
Now, if I hit API, I should get only A record with aid =1 as only it is associated with B.i starting with x.