10

次のnamedQueryがあります

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

しかし、私はエラーが発生しています

RIGHT_ROUND_BRACKET を予期していましたが、'(' が見つかりました

COALESCEnamedQueryでの処理方法は?

JPAでnull値を処理する他の方法はありますか?

4

3 に答える 3

11

合体はJPA 2.0 API でサポートされています。

コンストラクトは Hibernate 独自のnewものであり、必ずしもすべての JPA 実装でサポートされているわけではありません。最初に、オブジェクトを構築しようとせずにクエリを試してください。

select COALESCE(k.projectId,'N') as projectId, k.projectName from Emp o inner join o.projects k
于 2014-12-17T16:11:44.910 に答える
0

次の簡単な単体テストを試してみましたが、成功しました。

@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);
}

DepartmentEmployee クラス:

@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;
    }
}

従業員部門クラス:

@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;
    }
}

EclipseLink 2.5.0 を使用してテスト済み:

    <dependency>
        <groupId>org.eclipse.persistence</groupId>
        <artifactId>eclipselink</artifactId>
        <version>2.5.0</version>
    </dependency>
于 2014-12-22T13:26:23.707 に答える