JPA标准API将in表达式转换为多个OR

Question

带有“in”表达式的JPA标准API转换为多个“OR”而不是“in”

e. g.

我的主要模型

public class Person {
    ...

    @ManyToOne(fetch = FetchType.LAZY)
    @JoinColumn(name = "department_id")
    private Department department;

    ...
}

多对一关系

public class Department {
    @Id
    @Column(name="department_id")
    private Integer departmentId;

    @OneToMany(mappedBy="department")
    private List<Person> person;

    ...
}

在持久性中定义了这两个模型。xml（注意：它没有指定任何db方言）

<persistence xmlns="http://xmlns.jcp.org/xml/ns/persistence"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/persistence http://xmlns.jcp.org/xml/ns/persistence/persistence_2_1.xsd"
    version="2.1">

    <persistence-unit name="primary" transaction-type="JTA">

        <jta-data-source>java:openejb/Resource/jdbc/myDS</jta-data-source>

        <class>com.xyz.Person</class>
        <class>com.xyz.Department</class>
        ...

        <shared-cache-mode>ENABLE_SELECTIVE</shared-cache-mode>

        <properties>
            <property name="openjpa.Log" value="${open.jpa.log}" />
            <property name="openjpa.ConnectionFactoryProperties" value="printParameters=true" />
            <property name="openjpa.RuntimeUnenhancedClasses" value="unsupported" />
        </properties>
    </persistence-unit>

</persistence>

以下是用于构建查询的带条件API的代码：

CriteriaBuilder cb = em.getCriteriaBuilder();
CriteriaQuery<Person> criteriaQuery = cb.createQuery(Person.class);
Root<Person> personRoot = criteriaQuery.from(Person.class);

criteriaQuery.select(personRoot);

List<Predicate> predicateList = new ArrayList<>();
predicateList.add(personRoot.get(Person_.department).get(Department_.departmentId).in(Arrays.asList(1, 2, 3)));
// Using list because I actually need to add multiple conditions

criteriaQuery.where(predicateList.toArray(new Predicate[0]));

TypedQuery<Person> searchQuery = em.createQuery(criteriaQuery);

searchQuery.getResultList();

这将以以下格式打印查询：

SELECT t0.person_id, ... 
FROM person t0 
WHERE ((t0.department_id = ? OR t0.department_id = ? OR t0.department_id = ?) AND t0.department_id IS NOT NULL)

但我应该得到的是

SELECT t0.person_id, ... 
FROM person t0 
WHERE t0.department_id IN (?, ?, ?)

环境：

• 托梅7.0.3
• 嵌入式OpenJPA
• Mariadb

更新：

我也试着添加如下mariadb方言，但没有帮助

<property name="openjpa.jdbc.DBDictionary" value="mariadb" />

Answer 1

解决方案是使用表达式，例如：

List<Long> departmentIdsList = new ArrayList();
departmentIdsList.add(1L);
departmentIdsList.add(2L);
departmentIdsList.add(3L);
Expression<Long> exp = personRoot.get("departmentId");//"departmentId" field name to be equated
predicateList.add(exp.in(departmentIdsList));

上面的片段应该按照子句中的方式进行评估