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How to get collection of property in Join JPA criteria API

I'm trying to get a list of property policy_type_id from ListAttribute<Article, PolicyType>, but I can't figure out how to do it.
I come up with an inefficient method was select whole Collection of PolicyType then filter it later
Root<ArticleVersion> a = cq.from(ArticleVersion.class);
Join<ArticleVersion, Article> join1 = a.join(ArticleVersion_.article, JoinType.INNER);
cq.where(getCondition(cb, join1));
cq.multiselect(join1.get(Article_.article_id), join1.get(Article_.policyTypes), a);
Sadly, hibernate generate an error query like this
select article1_.article_id as col_0_0_, . as col_1_0_, articlever0_.article_version_id as col_2_0_ . As you can see, there is a . in select that make query broken (which I believe select all)
#Entity
#Table(name = "PolicyType", schema = "SM_Request")
#Getter
#Setter
#NoArgsConstructor
public class PolicyType {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
private int policy_type_id;
#ManyToMany(cascade = CascadeType.ALL, fetch = FetchType.LAZY)
#JoinTable(name = "PolicyTypeArticle", schema = "SM_Request",
joinColumns = #JoinColumn(name = "policy_type_id"), inverseJoinColumns = #JoinColumn(name = "article_id"))
#JsonIgnore
private List<Article> articles;
}

After long searching, I think that hibernate doesn't support query tuple of primitive types and list of objects (which is kinda sad, compare to LINQ to query). I decided to break down my query into smaller parts. First, I select tuples of article_id and ArticleVersion. After that, I select a list of PolicyType which also contains article_id, and union 2 lists back.
By the time I wrote this, I have an idea that I could select all 3 joins together and transform data the way I want. But It really depend on many aspects, like how many join or which type of join you're using, how fast data in each table grown (JOIN queries vs multiple queries)

HQL: How to query certain fields when one of those fields is a one-to-many List?

I am having difficulty writing a HQL query to select ONLY the caseid, title, and caseStatus fields from my Cases entity. The cases returned have to be distinct based on caseid. I do not want the name and userid fields to be included. I also do not want to use Lazy fetching for caseid, title, and caseStatus fields. Note that the caseStatus field is a one-to-many List. Below are the entities. The getters/setters are omitted to save space.
#Entity
#Table(name = "Cases")
public class Cases {
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
#Column(name = "caseid", nullable = false)
private Integer caseid;
private Integer userid;
private String name;
private String title;
#OrderBy("caseStatusId DESC")
#OneToMany(mappedBy = "cases", fetch = FetchType.EAGER)
private List<CaseStatus> caseStatus;
}
#Entity
#Table(name = "CaseStatus")
public class CaseStatus {
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
#Column(name = "caseStatusId", nullable = false)
private Integer caseStatusId;
private String info;
#ManyToOne(fetch = FetchType.LAZY)
#JoinColumn(name = "caseid")
private Cases cases;
}
My goal is to retrieve a distinct List<Cases> or List<Object[]> of the Cases entity containing only caseid, title, and a List<CaseStatus>. The List<CaseStatus> will contain CaseStatus objects with all of its fields populated.
public List<Object[]> getCases(String title) {
TypedQuery<Object[]> q = em.createQuery("select distinct c.caseid, c.title, cs "
+ "FROM Cases c join c.caseStatus cs "
+ "where c.title like :title", Object[].class);
q.setParameter("title", "%" + title + "%");
List<Object[]> results = q.getResultList();
return results;
}
The above method is close, but not correct because rather than returning a List<CaseStatus> in one of the indexes, it is only returning a single CaseStatus entity.
For example, if my DB contains a single Case with a List<CaseStatus> having a size of n for example, the results will be similar to the example below:
Example of results I'm getting now. Not correct:
List<Object[]> index 0:
Contains an Object[] where:
Object[0] = {some caseid}
Object[1] = {some title}
Object[2] = {1st CaseStatus}
List<Object[]> index 1:
Contains an Object[] where:
Object[0] = {same caseid as the one found in index 0 above}
Object[1] = {same title as the one found in index 0 above}
Object[2] = {2nd CaseStatus}
...
List<Object[]> index n-1:
Contains an Object[] where:
Object[0] = {same caseid as all the previous}
Object[1] = {same title as all the previous}
Object[2] = {nth CaseStatus}
Example of results I hope to achieve:
List<Object[]> index 0:
Contains an Object[] where:
Object[0] = {unique caseid}
Object[1] = {some title}
Object[2] = List<CaseStatus> with size of n
Updated the question. Instead of name, title, and List<CaseStatus>, the fields I want to retrieve are caseid, title, and List<CaseStatus>. caseid is the primary key of Cases.
I found various threads Select Collections with HQL - hibernate forum and Select collections with HQL - stackoverflow. It's pretty much the problem I ran into. Looks like no one found a solution in these threads.

Hibernates a bit confused about the query; in HQL do your join like this (apologies, I've not been able to test before posting due to wonky computer, but you should get the idea)
select distinct c from Cases c left join fetch c.caseStatus cs where....
the "fetch" makes it eager. Note that this will return an array of type Cases. You where clauses look about right.

In fact HQL is fully object-oriented and uses your classes structure in the Query, so by writing c.caseStatus HQL expects that your Cases class has a caseStatus property, which is wrong because it's a collection.
If you take a look at Hibernate HQL documentation you can see that:
Compared with SQL, however, HQL is fully object-oriented and understands notions like inheritance, polymorphism and association.
I think what you need to do here is to change your query so it matches your classes structures:
Query q = em.createQuery("select distinct c.name, c.title, cs.caseStatus FROM Cases c left join c.caseStatus where "
+ "c.name like :name and "
+ "c.title like :title");

Correct syntax should be
TypedQuery<Object[]> q = em.createQuery("select c.name, c.title, cs FROM Cases c "
+ "join c.caseStatus cs where "
+ "c.name = :name and "
+ "c.title = :title", Object[].class);
Return type will be List<Object[]>, where in first index of Object[] is c.name, second is c.title and third is associated caseStatus entity. It is possible to query for multiple instances (rows).
We need JOIN because relationship between CaseStatus and Case is mapped via collection.
SELECT cs
FROM Case c JOIN c.cases cs;

Why don't you just use
Query q = em.createQuery("select distinct c from Cases c where "
+ "c.name like :name and "
+ "c.title like :title");
Just try this. This may be a naive approach but should be able to solve the problem. You may be getting more fields than you required but the return type would be list of Cases.

spring data jpa unnecessary left join

I have the following model:
I want to get all Institutions (Intituciones) with specified sectorId.
In the tbInstitucion model I have a relationship with tbSector:
#ManyToOne(fetch=FetchType.LAZY)
#JoinColumn(name="`sectorId`")
private Sector sector;
is there a way to obtain a query like:
select *
from tbInstitucion
where sectorId = ?
I tried with: findBySector(Sector sector)
but with this I need an additional query to find the sector and findBySector is generating the following query:
select
generatedAlias0.institucionId,
generatedAlias0.institucionNombre
from
Institucion as generatedAlias0
left join
generatedAlias0.sector as generatedAlias1
where
generatedAlias1=:param0
tried with this other one:
findBySector_sectorId
which generates the above query as well.
Wouldn't be better to form a query like:
select *
from tbInstitucion
where sectorId = ?
Is there a way to get the above query?
Why is JPA generating the left join?

A quick review of the entity model
#Entity
class Institucion {
#ManyToOne(fetch=FetchType.LAZY)
#JoinColumn(name="`sectorId`")
private Sector sector;
}
is equivalent to:
#Entity
class Institucion {
#ManyToOne(cascade = {}
, fetch=FetchType.LAZY
, optional = true
, targetEntity = void.class)
#JoinColumn(columnDefinition = ""
, foreignKey = #ForeignKey
, insertable = true
, name="`sectorId`"
, nullable = true
, referencedColumnName = ""
, table = ""
, unique = false
, updatable = false)
private Sector sector;
}
Note #ManyToOne(optional = true) and #JoinColumn(nullable = true). This signifies to the ORM that the sector attribute of Institucion is optional and may not be set (to a non-null value) all the time.
How the entity model impacts repository queries
Now consider the following repository:
public interface InstitucionRepository extends CrudRepository<Institucion, Long> {
List<Institucion> findAllByInstitucionNombre(String nombre);
List<Institucion> findAllByInstitucionEmail(String email);
}
Given the entity declaration above, the repository methods should produce queries such as:
select
generatedAlias0
from
Institucion as generatedAlias0
left join
generatedAlias0.sector as generatedAlias1
where
generatedAlias0.institucionNombre=:param0
and
select
generatedAlias0
from
Institucion as generatedAlias0
left join
generatedAlias0.sector as generatedAlias1
where
generatedAlias0.institucionEmail=:param0
This is because the entity model indicates sector to be optional so the ORM needs to load Institucions without worrying about their sectors.
Following this pattern, the following repository method:
List<Institucion> findAllBySector(Sector sector);
translates to:
select
generatedAlias0
from
Institucion as generatedAlias0
left join
generatedAlias0.sector as generatedAlias1
where
generatedAlias1=:param0
Solution 1
If Institucion.sector is not optional, make it mandatory in the model too:
#ManyToOne(fetch=FetchType.LAZY, optional = false)
#JoinColumn(name="`sectorId`", nullable = false)
private Sector sector;
Solution 2
If Institucion.sector is indeed optional, only a manual query such as the one shown in #MaciejKowalski's answer will work.
Simplified query
The following query will also work:
List<Institucion> findAllBySectorSectorId(Long id);
This assumes that the model attribute names are exactly as shown in the post.

Left join is a default implicit joining strategy, also when using the #EntityGraph feature.
I would recommend using explicit #Query definition:
#Query("select i from institution i inner join i.sector s where s.id = :sectorId")
public Institution getBySector(#Param("sectorId") Integer sectorId);

HQL. Intersection of two lists

Update: look my answer below on how to check if 2 list intersects (both for #ElementCollection with string/enums and usual entities list mapped like #OneToMany)
I have an entity which contains #ElementCollectionfield with enums.
public enum StatusType {
NEW, PENDING, CLOSED;
}
#Entity
public class MyEntity {
#ElementCollection
#CollectionTable(name = "status_type", joinColumns = {#JoinColumn(name = "my_entity_id")})
#Column(name = "status_type", nullable = false)
private Set<StatusType > statusTypes = new HashSet<StatusType >();
...
}
Now I want to get all entities which contains status NEW or PENDING (or both).
I'm trying to use this query:
SELECT DISTINCT u FROM MyEntity u WHERE u.statusTypes in :statusTypes
But I'm getting exception: org.postgresql.util.PSQLException: No value specified for parameter 9.
How to properly query on collections and filter by intersections?

Problem solved by adding JOIN clause to HQL. Hibernate couldn't implicitly recognize that query needs JOIN clause. May be it will help someone:
SELECT DISTINCT u FROM MyEntity u
LEFT JOIN u.statusTypes statusTypes
WHERE statusTypes in :statusTypes
I set the query params like this:
query.setParameter( "statusTypes", listOfStatusTypesEnums);
It will select rows where at least one element of listOfStatusTypesEnums list is present in entity's statusTypes property (i.e. if 2 list are intersects in some way).
If you have usual list of entities (which are not #ElementCollection, but #OneToMany etc), same rule will work as well. Just use like this: LEFT JOIN u.subEntities subEntities WHERE subEntities.id in :subEntityIds

Order by count using Spring Data JpaRepository

I am using Spring Data JpaRepository and I find it extremely easy to use. I actually need all those features - paging, sorting, filtering. Unfortunately there is one little nasty thing that seems to force me to fall back to use of plain JPA.
I need to order by a size of associated collection. For instance I have:
#Entity
public class A{
#Id
private long id;
#OneToMany
private List<B> bes;
//boilerplate
}
and I have to sort by bes.size()
Is there a way to somehow customize the ordering still taking the advantage of pagination, filtering and other Spring Data great features?

I've solved the puzzle using hints and inspirations from:
Limiting resultset using #Query anotations by Koitoer
How to order by count() in JPA by MicSim
Exhaustive experiments on my own
The first and most important thing I've not been aware of about spring-data is that even using #Query custom methods one can still create paging queries by simply passing the Pageable object as parameter. This is something that could have been explicitely stated by spring-data documentation as it is definitely not obvious though very powerful feature.
Great, now the second problem - how do I actually sort the results by size of associated collection in JPA? I've managed to come to a following JPQL:
select new package.AwithBCount(count(b.id) as bCount,c) from A a join a.bes b group by a
where AwithBCount is a class that the query results are actually mapped to:
public class AwithBCount{
private Long bCount;
private A a;
public AwithBCount(Long bCount, A a){
this.bCount = bCount;
this.a = a;
}
//getters
}
Excited that I can now simply define my repository like the one below
public interface ARepository extends JpaRepository<A, Long> {
#Query(
value = "select new package.AwithBCount(count(b.id) as bCount,c) from A a join a.bes b group by a",
countQuery = "select count(a) from A a"
)
Page<AwithBCount> findAllWithBCount(Pageable pageable);
}
I hurried to try my solution out. Perfect - the page is returned but when I tried to sort by bCount I got disappointed. It turned out that since this is a ARepository (not AwithBCount repository) spring-data will try to look for a bCount property in A instead of AwithBCount. So finally I ended up with three custom methods:
public interface ARepository extends JpaRepository<A, Long> {
#Query(
value = "select new package.AwithBCount(count(b.id) as bCount,c) from A a join a.bes b group by a",
countQuery = "select count(a) from A a"
)
Page<AwithBCount> findAllWithBCount(Pageable pageable);
#Query(
value = "select new package.AwithBCount(count(b.id) as bCount,c) from A a join a.bes b group by a order by bCount asc",
countQuery = "select count(a) from A a"
)
Page<AwithBCount> findAllWithBCountOrderByCountAsc(Pageable pageable);
#Query(
value = "select new package.AwithBCount(count(b.id) as bCount,c) from A a join a.bes b group by a order by bCount desc",
countQuery = "select count(a) from A a"
)
Page<AwithBCount> findAllWithBCountOrderByCountDesc(Pageable pageable);
}
...and some additional conditional logic on service level (which could be probably encapsulated with an abstract repository implementation). So, although not extremely elegant, that made the trick - this way (having more complex entities) I can sort by other properties, do the filtering and pagination.

One option, which is much simpler than the original solution and which also has additional benefits, is to create a database view of aggregate data and link your Entity to this by means of a #SecondaryTable or #OneToOne.
For example:
create view a_summary_view as
select
a_id as id,
count(*) as b_count,
sum(value) as b_total,
max(some_date) as last_b_date
from b
Using #SecondaryTable
#Entity
#Table
#SecondaryTable(name = "a_summary_view",
pkJoinColumns = {#PrimaryKeyJoinColumn(name = "id", referencedColumnName= "id")})
public class A{
#Column(table = "a_summary_view")
private Integer bCount;
#Column(table = "a_summary_view")
private BigDecimal bTotal;
#Column(table = "a_summary_view")
private Date lastBDate;
}
You can now then sort, filer, query etc purely with reference to entity A.
As an additional advantage you have within your domain model data that may be expensive to compute in-memory e.g. the total value of all orders for a customer without having to load all orders or revert to a separate query.

Thank you #Alan Hay, this solution worked fine for me. I just had to set the foreignKey attribute of the #SecondaryTable annotation and everything worked fine (otherwise Spring Boot tried to add a foreignkey constraint to the id, which raise an error for a sql View).
Result:
#SecondaryTable(name = "product_view",
pkJoinColumns = {#PrimaryKeyJoinColumn(name = "id", referencedColumnName = "id")},
foreignKey = #javax.persistence.ForeignKey(ConstraintMode.NO_CONSTRAINT))

I don't know much about Spring Data but for JPQL, to sort the objects by size of associated collection, you can use the query
Select a from A a order by a.bes.size desc

You can use the name of an attribute found in the select clause as a sort property:
#Query(value = "select a, count(b) as besCount from A a join a.bes b group by a", countQuery = "select count(a) from A a")
Page<Tuple> findAllWithBesCount(Pageable pageable);
You can now sort on property besCount :
findAllWithBesCount(PageRequest.of(1, 10, Sort.Direction.ASC, "besCount"));

I used nativeQuery to arrange sorting by number of records from another table, pagable works.
#Query(value = "SELECT * FROM posts where posts.is_active = 1 and posts.moderation_status = 'ACCEPTED' " +
"group by posts.id order by (SELECT count(post_id) FROM post_comments where post_id = posts.id) desc",
countQuery = "SELECT count(*) FROM posts",
nativeQuery = true)
Page <Post> findPostsWithPagination(Pageable pageable);

For SpringBoot v2.6.6, accepted answer isn't working if you need to use pageable with child's side field especially when using #ManyToOne.
For the accepted answer:
You can return new object with static query method, which have to include order by count(b.id)
And also order by bCount isn't working.
Please use #AlanHay solution, it is working, but you can't use primitive field and change foreign key constraint. For instance, change long with Long. Because:
When saving a new entity Hibernate does think a record has to be written to the secondary table with a value of zero. (if you use primitive type)
Otherwise you will get an exception:
Caused by: org.postgresql.util.PSQLException: ERROR: cannot insert into view "....view"
Here is the example:
#Entity
#Table(name = "...")
#SecondaryTable(name = "a_summary_view,
pkJoinColumns = {#PrimaryKeyJoinColumn(name = "id",
referencedColumnName= "id")},
foreignKey = #javax.persistence.ForeignKey(name = "none"))
public class UserEntity {
#Id
private String id;
#NotEmpty
private String password;
#Column(table = "a_summary_view",
name = "b_count")
private Integer bCount;
}