Hibernate：如何将B映射为A的属性？ [英] Hibernate: how do I map one to one where B is a property of A?

问题描述

我有一个类A，拥有B类的属性。

A的SQL表对B没有任何了解。

B的SQL表包含A的外键。

如何映射（在hbm.xml中）以便B是属性A？我知道如何用Sets来做到这一点：

 < set name =btable =Bcascadeall-删除-孤儿> 
< key column =a_id> 
< composite-element class =B> 
< property name =bPropcolumn =b_proptype =string/> 
< / composite-element> 
< / set> 
  

事情是，B不是一组A.而只是一个元素。编辑：为了澄清一下，我的用例类似于 HTTP：//docs.jboss。 org / hibernate / orm / 3.6 / reference / en-US / html / associations.html＃assoc-unidirectional-m21 除外：

  create table Person（personId bigint not null primary key，addressId bigint not null）
 create table Address（addressId bigint不为null主键）
  

我有：

  create table Person（personId bigint不是null主键）
 create table地址（personId bigint，addressId bigint不为null主键）
  

<然而，我期望实际的地址类不包含对Person的任何引用：

  class Person {
地址地址; 
} 
 
 class地址{
 int id; 
} 
  

解决方案

此解决方案，但您可以通过使用几个包装方法解决此问题。为未映射的属性创建一个假的getter / setter对，为您提供所需的接口。因此：

  public class Person {
 private List< Address>地址; 
 //属性，实际获取者和设置者
 
 public Address getAddress（）{
 if（this.addresses == null || this.addresses.isEmpty（））{
返回null; 
} 
返回this.addresses.get（0）; 
} 
 $ b $ public void setAddress（Address address）{
 if（this.addresses == null）{
 this.addresses = new ArrayList< Address>（） ; 
} 
 this.addresses.clear（）; 
 this.addresses.add（address）; 
} 
} 
  

I have a class A, with property of class B.

The SQL table for A does not know anything about B.

The SQL table for B contains a foreign key to A.

How can I map (in hbm.xml) so that B is a property of A? I know how to do this with Sets:

<set name="b" table="B" cascade"all-delete-orphan">
    <key column="a_id">
    <composite-element class="B">
        <property name="bProp" column="b_prop" type="string"/>
    </composite-element>
</set>

Thing is, B is NOT a set of A. Rather just a single element. How might I go about mapping this?

---

Edit: To clarify a bit, my use case is similar to http://docs.jboss.org/hibernate/orm/3.6/reference/en-US/html/associations.html#assoc-unidirectional-m21 except instead of:

create table Person ( personId bigint not null primary key, addressId bigint not null )
create table Address ( addressId bigint not null primary key )

I have:

create table Person ( personId bigint not null primary key )
create table Address ( personId bigint, addressId bigint not null primary key )

However, I would expect then that the actual Address Class does not contain any references to Person:

class Person {
    Address address;
}

class Address {
    int id;
}

解决方案

I don't especially like this solution, but you can work around the issue by using a couple of wrapper methods. Create a fake getter/setter pair for an unmapped property that gives you the interface you need. Thus:

public class Person {
    private List<Address> addresses;
    // properties, real getters and setters

    public Address getAddress() {
        if (this.addresses == null || this.addresses.isEmpty()) {
            return null;
        }
        return this.addresses.get(0);
    }

    public void setAddress(Address address) {
        if (this.addresses == null) {
            this.addresses = new ArrayList<Address>();
        }
        this.addresses.clear();
        this.addresses.add(address);
    }
}

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