这是正确的（更好的）对象关系映射吗？ [英] Is this correct (better) object relational mapping?
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： Hibernate：太多行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
<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：
http://wiki.eclipse.org/EclipseLink/Examples/JPA/2.0/DerivedIdentifiers
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/DerivedIdentifiers
which 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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