JPA Criteria构建器IN子句查询

Question

如何为下面给出的JPQL查询编写条件生成器api查询？我在用JPA 2.2.

SELECT * 
FROM Employee e
WHERE e.Parent IN ('John','Raj')
ORDER BY e.Parent

Answer 1

这个标准设置应该可以解决问题:

CriteriaBuilder cb = entityManager.getCriteriaBuilder();
CriteriaQuery<Employee> q = cb.createQuery(Employee.class);


Root<Employee> root = q.from(Employee.class);
q.select(root);

List<String> parentList = Arrays.asList(new String[]{"John", "Raj"});

Expression<String> parentExpression = root.get(Employee_.Parent);
Predicate parentPredicate = parentExpression.in(parentList);
q.where(parentPredicate);
q.orderBy(cb.asc(root.get(Employee_.Parent));

q.getResultList();

我已经使用了重载CriteriaQuery.where此处方法,该方法接受Predicate..一个in在此情况下断言.

Answer 2

您也可以使用 Criteria API In 子句执行此操作，如下所示：

CriteriaBuilder cb = entityManager.getCriteriaBuilder();
CriteriaQuery<Employee> cq = cb.createQuery(Employee.class);
Root<Employee> root = cq.from(Employee.class);
List<String> parentList = Arrays.asList("John", "Raj");
In<String> in = cb.in(root.get(Employee_parent));
parentList.forEach(p -> in.value(p));

return entityManager
        .createQuery(cq.select(root)
        .where(in).orderBy(cb.asc(root.get(Employee_.Parent)))
        .getResultList();

查看我的Github以了解这个和几乎所有可能的标准示例。

Answer 3

List<String> parentList = Arrays.asList("John", "Raj");            
final CriteriaBuilder cb = entityManager.getCriteriaBuilder();
final CriteriaQuery<Employee> query = cb.createQuery(Employee.class);
final Root<Employee> employee = query.from(Employee.class);

query.select(employee).where(employee.get("Parent").in(parentList));

这应该可以正常工作。有关更多信息，请参阅 baeldung 的这篇文章。它非常足智多谋https://www.baeldung.com/jpa-criteria-api-in-expressions