Java JPA Map<String,String> 映射
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JPA Map<String,String> mapping
提问by Rafa de Castro
How can I map a Map in JPA without using Hibernate's classes?
如何在不使用 Hibernate 的类的情况下在 JPA 中映射 Map？
采纳答案by Chris K
Does not the following work for you?
以下不适合您吗？
@ManyToMany(cascade = CascadeType.ALL)
Map<String,EntityType> entitytMap = new HashMap<String, EntityType>();
EntityTypecould be any entity type, including a String.
EntityType可以是任何实体类型，包括String.
回答by Subhendu Mahanta
Suppose I have an entity named Book which is having a Map of chapters:
假设我有一个名为 Book 的实体，它有一个章节地图：
import java.io.Serializable;
import java.util.Map;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.JoinColumn;
import javax.persistence.JoinTable;
import org.hibernate.annotations.CollectionOfElements;
import org.hibernate.annotations.MapKey;
@Entity
public class Book implements Serializable{
@Column(name="BOOK_ID")
@Id
@GeneratedValue(strategy=GenerationType.AUTO)
private Long bookId;
@CollectionOfElements(targetElement=java.lang.String.class)
@JoinTable(name="BOOK_CHAPTER",
joinColumns=@JoinColumn(name="BOOK_ID"))
@MapKey (columns=@Column(name="CHAPTER_KEY"))
@Column(name="CHAPTER")
private Map<String,String> chapters;
public Long getBookId() {
return bookId;
}
public void setBookId(Long bookId) {
this.bookId = bookId;
}
public Map<String,String> getChapters() {
return chapters;
}
public void setChapters(Map<String,String> chapters) {
this.chapters = chapters;
}
}
It works for me.
这个对我有用。
回答by Jatin Sehgal
Although answer given by Subhendu Mahanta is correct. But @CollectionOfElementsis deprecated. You can use @ElementCollectioninstead:
尽管 Subhendu Mahanta 给出的答案是正确的。但@CollectionOfElements已弃用。您可以@ElementCollection改用：
@ElementCollection
@JoinTable(name="ATTRIBUTE_VALUE_RANGE", joinColumns=@JoinColumn(name="ID"))
@MapKeyColumn (name="RANGE_ID")
@Column(name="VALUE")
private Map<String, String> attributeValueRange = new HashMap<String, String>();
There is no need to create a separate Entity class for the Mapfield. It will be done automatically.
无需为该Map字段创建单独的实体类。它会自动完成。
回答by Taioli Francesco
A working example:
一个工作示例：
@ElementCollection(fetch=FetchType.EAGER)
@CollectionTable(name = "TABLENAME")
@MapKeyColumn(name = "KEY")
@Column(name = "VALUE")
public Map<String, String> getMap() {
return _map;
}
