JPA 2.0, Criteria API, Subqueries, In Expressions

问题

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?

回答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.

回答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();

回答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);

来源：https://stackoverflow.com/questions/4483576/jpa-2-0-criteria-api-subqueries-in-expressions

标签

jpa

subquery

jpa-2.0

criteria-api

in-subquery