many-to-many - spring Roo中如何做多对多关系，属性在de关系中？

Question

我一直在研究这个话题，还没有找到任何答案。我正在使用 spring roo，我想知道是否有一种方法可以与该关系中的属性建立多对多关系。例如我有两张表 Employee 和 MedicalEquipment，Employees 可以保留许多设备，设备可以被许多员工保留，但是我想存储这个保留发生的日期。

如果有人可以提供帮助，我将不胜感激。提前致谢!

Answer 1

• 为了实现与属性的多对多关系，您需要一个包含附加列的连接表。也就是说，除了 Employee 和 MedicalEquipment，您还需要第三个 EmployeeMedicalEquipment。

• 关于 JPA，您有两种选择:

  1. 将连接表映射到中间实体
  2. 将连接表映射到组件集合

前者更复杂，但它允许您进行双向导航(因为它是一个实体，因此可以共享引用)，后者在构建和使用它时更简单，但你不能使用它在实体之间导航(但是，您可以编写查询来检索所需的对象)

• Roo 是一个代码生成器，但它缺少 JPA 提供的所有选项，因此您必须编辑 Java 类和测试类。 View 的构建也有一些限制，因此您也需要对其进行编辑。

在我的例子中，我需要创建一个中间实体，因为该表属于一个已经存在的遗留数据库。

我做了这样的事情:

entity --class ~.domain.Employee --table T_Employee [etc...]
field [etc...]

entity --class ~.domain.MedicalEquipment --table T_MedicalEquipment [etc...]
field [etc...]

entity --class ~.domain.EmployeeMedicalEquipment --table T_Employee_MedicalEquipment --identifierType ~.domain.EmployeeMedicalEquipmentId
//to store the date this reserve took place.
field date --fieldName reserveDate --column C_reserveDate [etc...]

//Bidirectional: you´ll need @JoinColumn insertable = false and updatable = false
field reference --fieldName employee --type ~.domain.Employee --cardinality MANY_TO_ONE 

//Bidirectional: you'll need @JoinColumn insertable = false and updatable = false
field reference --fieldName medicalEquipment --type ~.MedicalEquipment --cardinality MANY_TO_ONE 

//Join table's composite primary key
field string --fieldName employeeId --column employee_ID --class ~.domain.EmployeeMedicalEquipmentId [etc...]
field string --fieldName medicalEquipmentId --column medicalEquipment_ID  --class ~.domain.EmployeeMedicalEquipmentId [etc...]

//Now, it's time to complete the relationship:
focus --class ~.domain.Employee
field set --type ~.domain.EmployeeMedicalEquipment --fieldName medicalEquipments --cardinality ONE_TO_MANY --mappedBy employee
focus --class ~.domain.MedicalEquipment
field set --type ~.domain.EmployeeMedicalEquipment --fieldName employees --cardinality ONE_TO_MANY --mappedBy medicalEquipment

此外，您需要通过使用构造函数管理关联两侧的集合来保证引用完整性。所以你需要以这种方式编辑类:

@RooEntity(...
public class EmployeeMedicalEquipment {

@ManyToOne
@JoinColumn(name = "employeeId", referencedColumnName = "employeeId", insertable = false, updatable = false)
private Employee employee;

@ManyToOne
@JoinColumn(name="medicalEquipmentId", referencedColumnName="medicalEquipmentId", insertable=false, updatable=false)
private MedicalEquipment medicalEquipment;

/**
 * No-arg constructor for JavaBean tools
 */
public EmployeeMedicalEquipment() {
}


/**
 * Full constructor, the Employee and MedicalEquipment instances have to have an identifier value, they have to be in detached or persistent state.
 * This constructor takes care of the bidirectional relationship by adding the new instance to the collections on either side of the
 * many-to-many association (added to the collections)
 */
public EmployeeMedicalEquipment(Employee employee, MedicalEquipment medicalEquipment, Date reserveDate) {
    this.setReserveDate (reserveDate);

    this.employee = employee;
    this.medicalEquipment = medicalEquipment;

    this.setId(new EmployeeMedicalEquipmentId(employee.getId(),  medicalEquipment.getId());

    // If Employee or MedicalEquipment  Guarantee referential integrity
    employee.getMedicalEquipments().add(this);
    medicalEquipment.getEmployees().add(this);
}
...

}

我试图给你一个 Roo 配置的例子。

您可以在 Manning “Java Persistence with Hibernate”一书中的第 7.2.3 章中找到对 JPA 内容的更好解释。

注意:如果使用 roo 1.2.1，count 查询会生成带有“count(id1, id2)”的 SQL，并非所有数据库都支持，包括 HSQLDB。您可以像这样自定义它:

...

-@RooJpaActiveRecord(identifierType = EmployeeMedicalEquipmentId.class, table = "T_Employee_MedicalEquipment")
+@RooJpaActiveRecord(identifierType = EmployeeMedicalEquipmentId.class, table = "T_Employee_MedicalEquipment", countMethod="")
...
public class EmployeeMedicalEquipment {
...
    // count method initially copied from ActiveRecord aspect
    public static long countEmployeeMedicalEquipments() {
-       return entityManager().createQuery("SELECT COUNT(o) FROM EmployeeMedicalEquipment o", Long.class).getSingleResult();
+       return entityManager().createQuery("SELECT COUNT(o.employee) FROM EmployeeMedicalEquipment o", Long.class).getSingleResult();
        }
    }