COALESCE in JPA namedQuery

Question

我有以下命名

查询

select new test.entity.Emp(COALESCE(k.projectId,'N')
as projectId, k.projectName) from Emp o inner join o.projects k 

但是我收到错误

期待RIGHT_ROUND_BRACKET，发现"(">

如何处理命名查询中的COALESCE？

还有其他方法可以在 JPA 中处理空值吗？

Answer 1

Coalesce 由 JPA 2.0 API 支持。

new结构是Hibernate专有的，不一定在所有JPA实现中都支持。首先尝试查询，而不尝试构造对象：

select COALESCE(k.projectId,'N') as projectId, k.projectName from Emp o inner join o.projects k

Answer 2

你的括号被弄乱了，或者你有一个多余的别名子句，当你正确地缩进你的语句时，它变得很容易看到。

select 
    new test.entity.Emp(
        COALESCE(k.projectId,'N') as projectId, 
        k.projectName
    ) 
from Emp o inner join o.projects k 

试试这个：

select 
    new test.entity.Emp(
        COALESCE(k.projectId,'N'), 
        k.projectName
    ) 
from Emp o inner join o.projects k 

Answer 3

我尝试了以下简单的单元测试，成功通过：

@Test
public void coalesceTest() {
    EntityManagerFactory entityManagerFactory = Persistence.createEntityManagerFactory("PersistenceUnit");
    EntityManager entityManager = entityManagerFactory.createEntityManager();
    EntityTransaction transaction = entityManager.getTransaction();
    DepartmentEmployee employee = new DepartmentEmployee();
    EmployeeDepartment department = new EmployeeDepartment();
    department.getEmployees().add(employee);
    employee.setDepartment(department);
    transaction.begin();
    try {
        entityManager.persist(employee);
        entityManager.persist(department);
        transaction.commit();
        Assert.assertTrue("Employee not persisted", employee.getId() > 0);
        Assert.assertTrue("Department not persisted", department.getId() > 0);
    } catch (Exception x) {
        if(transaction.isActive()) {
            transaction.rollback();
        }
        Assert.fail("Failed to persist: " + x.getMessage());
    }
    TypedQuery<String> query = entityManager.createQuery("select coalesce(e.name, 'No Name') from EmployeeDepartment d join d.employees e", String.class);
    String employeeName = query.getSingleResult();
    Assert.assertEquals("Unexpected query result", "No Name", employeeName);
}

部门员工类别：

@Entity
public class DepartmentEmployee implements Serializable {
    @Id
    @GeneratedValue
    private int id;
    private String name;
    @ManyToOne
    private EmployeeDepartment department;
    public int getId() {
        return id;
    }
    public String getName() {
        return name;
    }
    public void setName(String name) {
        this.name = name;
    }
    public EmployeeDepartment getDepartment() {
        return department;
    }
    public void setDepartment(EmployeeDepartment department) {
        this.department = department;
    }
}

员工部门类：

@Entity
public class EmployeeDepartment implements Serializable {
    @Id
    @GeneratedValue
    private int id;
    @OneToMany
    private List<DepartmentEmployee> employees;
    public EmployeeDepartment() {
        employees = new ArrayList<DepartmentEmployee>();
    }
    public int getId() {
        return id;
    }
    public List<DepartmentEmployee> getEmployees() {
        return employees;
    }
    public void setEmployees(List<DepartmentEmployee> employees) {
        this.employees = employees;
    }
}

使用 EclipseLink 2.5.0 进行测试：

    <dependency>
        <groupId>org.eclipse.persistence</groupId>
        <artifactId>eclipselink</artifactId>
        <version>2.5.0</version>
    </dependency>