JPA 2.0, Criteria API, Subqueries, In Expressions - jpa

Question

I have tried to write a query statement with a subquery and an IN expression for many times. But I have never succeeded.
I always get the exception, " Syntax error near keyword 'IN' ", the query statement was build like this,
SELECT t0.ID, t0.NAME
FROM EMPLOYEE t0
WHERE IN (SELECT ?
FROM PROJECT t2, EMPLOYEE t1
WHERE ((t2.NAME = ?) AND (t1.ID = t2.project)))
I know the word before 'IN' lose.
Have you ever written such a query? Any suggestion?

Answer 1

Below is the pseudo-code for using sub-query using Criteria API.
CriteriaBuilder criteriaBuilder = entityManager.getCriteriaBuilder();
CriteriaQuery<Object> criteriaQuery = criteriaBuilder.createQuery();
Root<EMPLOYEE> from = criteriaQuery.from(EMPLOYEE.class);
Path<Object> path = from.get("compare_field"); // field to map with sub-query
from.fetch("name");
from.fetch("id");
CriteriaQuery<Object> select = criteriaQuery.select(from);
Subquery<PROJECT> subquery = criteriaQuery.subquery(PROJECT.class);
Root fromProject = subquery.from(PROJECT.class);
subquery.select(fromProject.get("requiredColumnName")); // field to map with main-query
subquery.where(criteriaBuilder.and(criteriaBuilder.equal("name",name_value),criteriaBuilder.equal("id",id_value)));
select.where(criteriaBuilder.in(path).value(subquery));
TypedQuery<Object> typedQuery = entityManager.createQuery(select);
List<Object> resultList = typedQuery.getResultList();
Also it definitely needs some modification as I have tried to map it according to your query. Here is a link http://www.ibm.com/developerworks/java/library/j-typesafejpa/ which explains concept nicely.

Answer 2

Late resurrection.
Your query seems very similar to the one at page 259 of the book Pro JPA 2:
Mastering the Java Persistence API, which in JPQL reads:
SELECT e
FROM Employee e
WHERE e IN (SELECT emp
FROM Project p JOIN p.employees emp
WHERE p.name = :project)
Using EclipseLink + H2 database, I couldn't get neither the book's JPQL nor the respective criteria working. For this particular problem I have found that if you reference the id directly instead of letting the persistence provider figure it out everything works as expected:
SELECT e
FROM Employee e
WHERE e.id IN (SELECT emp.id
FROM Project p JOIN p.employees emp
WHERE p.name = :project)
Finally, in order to address your question, here is an equivalent strongly typed criteria query that works:
CriteriaBuilder cb = em.getCriteriaBuilder();
CriteriaQuery<Employee> c = cb.createQuery(Employee.class);
Root<Employee> emp = c.from(Employee.class);
Subquery<Integer> sq = c.subquery(Integer.class);
Root<Project> project = sq.from(Project.class);
Join<Project, Employee> sqEmp = project.join(Project_.employees);
sq.select(sqEmp.get(Employee_.id)).where(
cb.equal(project.get(Project_.name),
cb.parameter(String.class, "project")));
c.select(emp).where(
cb.in(emp.get(Employee_.id)).value(sq));
TypedQuery<Employee> q = em.createQuery(c);
q.setParameter("project", projectName); // projectName is a String
List<Employee> employees = q.getResultList();

Answer 3

CriteriaBuilder criteriaBuilder = em.getCriteriaBuilder();
CriteriaQuery<Employee> criteriaQuery = criteriaBuilder.createQuery(Employee.class);
Root<Employee> empleoyeeRoot = criteriaQuery.from(Employee.class);
Subquery<Project> projectSubquery = criteriaQuery.subquery(Project.class);
Root<Project> projectRoot = projectSubquery.from(Project.class);
projectSubquery.select(projectRoot);
Expression<String> stringExpression = empleoyeeRoot.get(Employee_.ID);
Predicate predicateIn = stringExpression.in(projectSubquery);
criteriaQuery.select(criteriaBuilder.count(empleoyeeRoot)).where(predicateIn);

Answer 4

You can use double join, if table A B are connected only by table AB.
public static Specification<A> findB(String input) {
return (Specification<A>) (root, cq, cb) -> {
Join<A,AB> AjoinAB = root.joinList(A_.AB_LIST,JoinType.LEFT);
Join<AB,B> ABjoinB = AjoinAB.join(AB_.B,JoinType.LEFT);
return cb.equal(ABjoinB.get(B_.NAME),input);
};
}
That's just an another option
Sorry for that timing but I have came across this question and I also wanted to make SELECT IN but I didn't even thought about double join.
I hope it will help someone.