JPA 2.0,Criteria API,子查询,表达式

Question

我曾尝试多次使用子查询和IN表达式编写查询语句.但我从来没有成功过.

我总是得到异常,"关键字'IN'附近的语法错误",查询语句是这样构建的,

SELECT t0.ID, t0.NAME
FROM EMPLOYEE t0
WHERE IN (SELECT ?
          FROM PROJECT t2, EMPLOYEE t1
          WHERE ((t2.NAME = ?) AND (t1.ID = t2.project)))

我知道'IN'失败前的这个词.

你有没有写过这样的问题？有什么建议吗？

Answer 1

下面是使用Criteria API使用子查询的伪代码.

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();

它也肯定需要一些修改,因为我试图根据您的查询映射它.这是一个链接http://www.ibm.com/developerworks/java/library/j-typesafejpa/,很好地解释了概念.

Answer 2

晚复活.

您的查询看起来非常类似于Pro JPA 2:掌握Java持久性API一书的第259页,其在JPQL中读取:

SELECT e 
FROM Employee e 
WHERE e IN (SELECT emp
              FROM Project p JOIN p.employees emp 
             WHERE p.name = :project)

使用EclipseLink + H2数据库,我无法获得本书的JPQL和相应的标准.对于这个特殊的问题,我发现如果你直接引用id而不是让持久性提供程序找出它,一切都按预期工作:

SELECT e 
FROM Employee e 
WHERE e.id IN (SELECT emp.id
                 FROM Project p JOIN p.employees emp 
                WHERE p.name = :project)

最后,为了解决您的问题,这里有一个等效的强类型条件查询,它有效:

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);