我曾多次尝试编写带有子查询和IN
表达式的查询语句。但我从来没有成功过。
我总是得到一个异常,“Syntax error near keyword 'IN‘",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‘lose之前的那个词。
你写过这样的查询吗?有什么建议吗?
发布于 2011-01-12 19:02:04
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();
此外,它肯定需要一些修改,因为我已经尝试根据您的查询映射它。这里有一个链接http://www.ibm.com/developerworks/java/library/j-typesafejpa/,它很好地解释了概念。
发布于 2012-04-03 04:09:06
晚间复活。
您的查询看起来与Pro JPA 2: Mastering the Java Persistence API一书第259页的查询非常相似,在JPQL语言中是这样写的:
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,我既无法获得该书的JPQL,也无法获得相应的标准。对于这个特定的问题,我发现如果你直接引用id,而不是让持久化提供者来解决,那么一切都会按预期进行:
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,下面是一个等效的强类型条件查询:
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();
发布于 2018-09-06 03:20:47
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);
https://stackoverflow.com/questions/4483576
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