这是正确的（更好的）对象关系映射吗？

问题描述

我真的被困在我的项目中使用Hibernate。我在这里描述了一些问题：我开始想知道我的代码是否正确。我正在研究一个巨大的项目，我必须自己定义带有注释的映射类。但是当问题开始出现时，我决定重新创建与项目分开的数据库的一部分，并尝试在IDE中生成实体。

我有两个表：我和选项。 我的有主键：列 qwerty 和属性。 属性是 Option 表中的外键。当然 Option 有属性作为主键。

<在我的解决方案中，我创建了 @Embeddable MyPK 类，它有两个属性： String qwerty 和字符串属性。在我的 My 实体中，我当然有 @EmbeddedId MyPK 和属性（与 MyPK 中的列名相同），但是这是 Option 对象，不是字符串，如 MyPK 。

  @ManyToOne 
 @JoinColumn（name =property，nullable = false，insertable = false，updatable = false）
保护选项选项; 
  

这是由Intellij Idea中的Hibernate Tools生成的实体。没有 EmbeddedId ，但有 @IdClass 。我认为 @IdClass 仅适用于基本类型。但我有一个对象作为我的主键的一部分。但是在这里也有 OptionEntity 对象。这是正确的保持一列的基本类型和对象类型？

  @ javax.persistence.IdClass（test.go.MyEntityPK .class）
 @ javax.persistence.Table（name =MY，schema =PUBLIC，catalog =PUBLIC）
 @Entity 
 public class MyEntity {
私人字符串qwerty; 
 
 @ javax.persistence.Column（name =QWERTY）
 @Id 
 public String getQwerty（）{
 return qwerty; 
} 
 
 public void setQwerty（String qwerty）{
 this.qwerty = qwerty; 
} 
 
私人字符串文本; 
 
 @ javax.persistence.Column（name =TEXT）
 @Basic 
 public String getText（）{
 return text; 
} 
 
 public void setText（String text）{
 this.text = text; 
} 
 
私人字符串lang; 
 
 @ javax.persistence.Column（name =PROPERTY）
 @Id 
 public String getProperty（）{
 return property; 
} 
 
 public void setProperty（String property）{
 this.property = property; 
} 
 
 @Override 
 public boolean equals（Object o）{
 // equals 
} 
 
 @Override 
 public int hashCode（）{
 // hashCode 
} 
 
 private OptionEntity optionByProperty; 
 
 @ManyToOne 
 @ javax.persistence.JoinColumn（name =PROPERTY，referencedColumnName =PROPERTY，nullable = false）
 public OptionEntity getOptionByProperty（）{
返回optionByProperty; 
} 
 
 public void setOptionByProperty（OptionEntity optionByProperty）{
 this.optionByProperty = optionByProperty; 
 
 
 $ / code $ / pre 
 $ b 
这是 MyEntityPK  code>生成的类：
  public class MyEntityPK implements Serializable {
 private String qwerty; 
 
 @Id 
 @Column（name =qwerty）
 public String getQwerty（）{
 return qwerty; 
} 
 
 public void setQwerty（String qwerty）{
 this.qwerty = qwerty; 
} 
 
私有字符串属性; 
 
 @Id 
 @Column（name =PROPERTY）
 public String getProperty（）{
 return property; 
} 
 
 public void setProperty（String property）{
 this.property = property; 
} 
 
 @Override 
 public boolean equals（Object o）{
 // equals 
} 
 
 @Override 
 public int hashCode（）{
 // hashCode 
} 
} 
  
  OptionEntity 下方。这个实体没有特别的要点。我只希望在版本属性上 @Version 注释，并且 List< MyEntity> 而不是集合< MyEntity> 。
  @ javax.persistence.Table（name =OPTION，schema =PUBLIC，catalog =PUBLIC）
 @Entity 
 public class OptionEntity {
 private Long version; 
 
 @ javax.persistence.Column（name =VERSION）
 @Basic 
 public Long getVersion（）{
 return version; 
} 
 
 public void setVersion（Long version）{
 this.version = version; 
} 
 
私有字符串属性; 
 
 @ javax.persistence.Column（name =PROPERTY）
 @Id 
 public String getProperty（）{
 return property; 
} 
 
 public void setProperty（String property）{
 this.property = property; 
} 
 
 @Override 
 public boolean equals（Object o）{
 // equals 
} 
 
 @Override 
 public int hashCode（）{
 // hashcode 
} 
 
 private Collection< MyEntity> myByProperty; 
 
 @OneToMany（mappedBy =optionByProperty）
 public Collection< MyEntity> getMyByProperty（）{
返回myByProperty; 
} 
 
 public void setMyByProperty（Collection< MyEntity> myByProperty）{
 this.myByProperty = myByProperty; 
} 
} 
  
什么样的选择是最合适的问题？我已经描述过的或粘贴的那个？
解决方案
查看使用派生ID的JPA 2.0示例：
  
这似乎是你以后的样子。 
 
 
 
在JPA中，您不需要为您的ID使用embeddedId，但是如果使用组合PK，您确实需要一个类来保存组成PK的多个值。这个类的实例被传递给em.find方法，它可以是EmbeddedId或PKClass。我更喜欢自己使用PKClass，但它取决于您 - 使用嵌入式只是将字段放置在可嵌入类中，因此您可以使用嵌入式对象来设置映射并访问值。如果使用pkClass，则不需要在其中注释的字段/属性，因为它们是直接在实体内访问和映射的。  
I have really got stuck with using Hibernate in my project. Some issues I have described here:  Hibernate: getting too many rows I have started wondering if my code is correct. I am working on a huge project and I have had to define mapping classes with annotations on my own. But when the problems have began to occur I have decided to recreate part of database separate to the project and try to generate entities in IDE.
I have two tables: My and Option. My has primary key: column qwerty and property. Propertyis the foreign key from Option table. And of course Option has property as a primary key.
In my solution I have created @Embeddable MyPK class with two properties: String qwerty and String property. In my My entity I have of course @EmbeddedId MyPK and also property (the same column name as in the MyPK) but is this Option object, not String as in the MyPK.
@ManyToOne
@JoinColumn(name = "property", nullable = false, insertable = false, updatable = false)
protected Option option;
This is entity generated by Hibernate Tools in Intellij Idea. There isn't EmbeddedId, but there is @IdClass. I have thought that @IdClass is only for basic types. But I have a object as a part of my primary key. However there is also OptionEntity object here. Is this correct to keep basic type and object type for one column?
@javax.persistence.IdClass(test.go.MyEntityPK.class)
@javax.persistence.Table(name = "MY", schema = "PUBLIC", catalog = "PUBLIC")
@Entity
public class MyEntity {
    private String qwerty;

    @javax.persistence.Column(name = "QWERTY")
    @Id
    public String getQwerty() {
        return qwerty;
    }

    public void setQwerty(String qwerty) {
        this.qwerty = qwerty;
    }

    private String text;

    @javax.persistence.Column(name = "TEXT")
    @Basic
    public String getText() {
        return text;
    }

    public void setText(String text) {
        this.text = text;
    }

    private String lang;

    @javax.persistence.Column(name = "PROPERTY")
    @Id
    public String getProperty() {
        return property;
    }

    public void setProperty(String property) {
        this.property= property;
    }

    @Override
    public boolean equals(Object o) {
        //equals
    }

    @Override
    public int hashCode() {
        //hashCode
    }

    private OptionEntity optionByProperty;

    @ManyToOne
    @javax.persistence.JoinColumn(name = "PROPERTY", referencedColumnName = "PROPERTY", nullable = false)
    public OptionEntity getOptionByProperty() {
        return optionByProperty;
    }

    public void setOptionByProperty(OptionEntity optionByProperty) {
        this.optionByProperty = optionByProperty;
    }
}
This is MyEntityPK generated class:
public class MyEntityPK implements Serializable {
    private String qwerty;

    @Id
    @Column(name = "qwerty")
    public String getQwerty() {
        return qwerty;
    }

    public void setQwerty(String qwerty) {
        this.qwerty = qwerty;
    }

    private String property;

    @Id
    @Column(name = "PROPERTY")
    public String getProperty() {
        return property;
    }

    public void setProperty(String property) {
        this.property = property;
    }

    @Override
    public boolean equals(Object o) {
        //equals
    }

    @Override
    public int hashCode() {
        //hashCode
    }
}
OptionEntity below. No special points aren't in this entity. I would like only @Version annotation on version property and also List<MyEntity> instead of Collection<MyEntity>.
@javax.persistence.Table(name = "OPTION", schema = "PUBLIC", catalog = "PUBLIC")
@Entity
public class OptionEntity {
    private Long version;

    @javax.persistence.Column(name = "VERSION")
    @Basic
    public Long getVersion() {
        return version;
    }

    public void setVersion(Long version) {
        this.version = version;
    }

    private String property;

    @javax.persistence.Column(name = "PROPERTY")
    @Id
    public String getProperty() {
        return property;
    }

    public void setProperty(String property) {
        this.property = property;
    }

    @Override
    public boolean equals(Object o) {
        //equals
    }

    @Override
    public int hashCode() {
        //hashcode
    }

    private Collection<MyEntity> myByProperty;

    @OneToMany(mappedBy = "optionByProperty")
    public Collection<MyEntity> getMyByProperty() {
        return myByProperty;
    }

    public void setMyByProperty(Collection<MyEntity> myByProperty) {
        this.myByProperty = myByProperty;
    }
}
What option is the most proper and less problematic? The one that I have described or the one that pasted?
 解决方案 
Check out JPA 2.0 examples using derived Ids:http://wiki.eclipse.org/EclipseLink/Examples/JPA/2.0/DerivedIdentifierswhich seem to be what you are after.   
In JPA, you do not need an embeddedId for your ids, but if using a composite PK, you do need a class to hold the multiple values that make up the pk.  Instances of this class are passed to the em.find method, and it can be either an EmbeddedId or a PKClass.  I prefer using PKClass myself, but its up to you - using an embedded just places the fields within the embeddedable class, so you use the embedded object to set the mappings and access the values.  If using a pkClass, you do not need the fields/properties annotated within it since they are accessed and mapped within the entity directly.  
                        
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