获得具有持久List类型字段的实体的最聪明的方法是什么?

Command.java

package persistlistofstring;

import java.io.Serializable;
import java.util.ArrayList;
import java.util.List;
import javax.persistence.Basic;
import javax.persistence.Entity;
import javax.persistence.EntityManager;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.Persistence;

@Entity
public class Command implements Serializable {

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    Long id;
    @Basic
    List<String> arguments = new ArrayList<String>();

    public static void main(String[] args) {
        Command command = new Command();

        EntityManager em = Persistence
                .createEntityManagerFactory("pu")
                .createEntityManager();
        em.getTransaction().begin();
        em.persist(command);
        em.getTransaction().commit();
        em.close();

        System.out.println("Persisted with id=" + command.id);
    }
}

这段代码产生:

> Exception in thread "main" javax.persistence.PersistenceException: No Persistence provider for EntityManager named pu: Provider named oracle.toplink.essentials.PersistenceProvider threw unexpected exception at create EntityManagerFactory: 
> oracle.toplink.essentials.exceptions.PersistenceUnitLoadingException
> Local Exception Stack: 
> Exception [TOPLINK-30005] (Oracle TopLink Essentials - 2.0.1 (Build b09d-fcs (12/06/2007))): oracle.toplink.essentials.exceptions.PersistenceUnitLoadingException
> Exception Description: An exception was thrown while searching for persistence archives with ClassLoader: sun.misc.Launcher$AppClassLoader@11b86e7
> Internal Exception: javax.persistence.PersistenceException: Exception [TOPLINK-28018] (Oracle TopLink Essentials - 2.0.1 (Build b09d-fcs (12/06/2007))): oracle.toplink.essentials.exceptions.EntityManagerSetupException
> Exception Description: predeploy for PersistenceUnit [pu] failed.
> Internal Exception: Exception [TOPLINK-7155] (Oracle TopLink Essentials - 2.0.1 (Build b09d-fcs (12/06/2007))): oracle.toplink.essentials.exceptions.ValidationException
> Exception Description: The type [interface java.util.List] for the attribute [arguments] on the entity class [class persistlistofstring.Command] is not a valid type for a serialized mapping. The attribute type must implement the Serializable interface.
>         at oracle.toplink.essentials.exceptions.PersistenceUnitLoadingException.exceptionSearchingForPersistenceResources(PersistenceUnitLoadingException.java:143)
>         at oracle.toplink.essentials.ejb.cmp3.EntityManagerFactoryProvider.createEntityManagerFactory(EntityManagerFactoryProvider.java:169)
>         at javax.persistence.Persistence.createEntityManagerFactory(Persistence.java:110)
>         at javax.persistence.Persistence.createEntityManagerFactory(Persistence.java:83)
>         at persistlistofstring.Command.main(Command.java:30)
> Caused by: 
> ...

当前回答

当使用JPA的Hibernate实现时,我发现只需将类型声明为ArrayList而不是List就可以允许Hibernate存储数据列表。

显然,与创建Entity对象列表相比,这有许多缺点。没有惰性加载,不能从其他对象引用列表中的实体,可能在构造数据库查询时更加困难。然而,当您处理相当原始类型的列表时,您总是希望与实体一起急切地获取,那么这种方法对我来说似乎很好。

@Entity
public class Command implements Serializable {

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    Long id;

    ArrayList<String> arguments = new ArrayList<String>();


}

其他回答

根据Java持久化与Hibernate

用注释映射值类型集合[…]。在撰写本文时,它还不是Java持久性标准的一部分

如果你正在使用Hibernate,你可以这样做:

@CollectionOfElements(targetElement = String.class)
@JoinTable(name = "foo", joinColumns = @JoinColumn(name = "foo_id"))
@IndexColumn(name = "POSITION", base = 1)
@Column(name = "baz", nullable = false)
private List<String> arguments = new ArrayList<String>();

更新:注意,这个功能现在在JPA2中可用。

下面是使用@Converter和StringTokenizer存储Set的解决方案。再检查一下@jonck-van-der-kogel的解决方案。

在你的实体类中:

@Convert(converter = StringSetConverter.class)
@Column
private Set<String> washSaleTickers;

StringSetConverter:

package com.model.domain.converters;

import javax.persistence.AttributeConverter;
import javax.persistence.Converter;
import java.util.HashSet;
import java.util.Set;
import java.util.StringTokenizer;

@Converter
public class StringSetConverter implements AttributeConverter<Set<String>, String> {
    private final String GROUP_DELIMITER = "=IWILLNEVERHAPPEN=";

    @Override
    public String convertToDatabaseColumn(Set<String> stringList) {
        if (stringList == null) {
            return new String();
        }
        return String.join(GROUP_DELIMITER, stringList);
    }

    @Override
    public Set<String> convertToEntityAttribute(String string) {
        Set<String> resultingSet = new HashSet<>();
        StringTokenizer st = new StringTokenizer(string, GROUP_DELIMITER);
        while (st.hasMoreTokens())
            resultingSet.add(st.nextToken());
        return resultingSet;
    }
}

我想要的是在一个表列中持久化一组string的简单方法。

我最终使用JSON,因为MySQL 5.7+有原生支持。 这是我的解决方案

    @Column(name = "eligible_approvers", columnDefinition = "json")
    @Convert(converter = ArrayJsonConverter.class)
    private Set<String> eligibleApprovers;

然后写一个很基本的转换器

@Converter(autoApply = true)
public class ArrayJsonConverter implements AttributeConverter<Set, String> {

    static final ObjectMapper mapper = new ObjectMapper();

    @Override
    public String convertToDatabaseColumn(Set list) {
        if (list == null)
            return null;
        try {
            return mapper.writeValueAsString(list);
        } catch (JsonProcessingException e) {
            throw new RuntimeException(e);
        }
    }


    @Override
    public Set convertToEntityAttribute(String dbJson) {
        if (dbJson == null)
            return null;
        try {
            return mapper.readValue(dbJson, new TypeReference<Set<String>>() {
            });
        } catch (JsonProcessingException e) {
            throw new RuntimeException(e);
        }
    }
}

这个答案是在JPA2之前实现的,如果你正在使用JPA2,请参阅上面的ElementCollection答案:

模型对象内的对象列表通常被认为是与另一个对象的“OneToMany”关系。但是,String本身并不是一对多关系的允许客户端,因为它没有ID。

因此,您应该将字符串列表转换为包含ID和String的参数类JPA对象列表。您可以潜在地使用String作为ID,这将在您的表中节省一点空间,既可以删除ID字段,也可以合并字符串相等的行,但您将失去将参数按原始顺序排序的能力(因为您没有存储任何排序信息)。

Alternatively, you could convert your list to @Transient and add another field (argStorage) to your class that is either a VARCHAR() or a CLOB. You'll then need to add 3 functions: 2 of them are the same and should convert your list of Strings into a single String (in argStorage) delimited in a fashion that you can easily separate them. Annotate these two functions (that each do the same thing) with @PrePersist and @PreUpdate. Finally, add the third function that splits the argStorage into the list of Strings again and annotate it @PostLoad. This will keep your CLOB updated with the strings whenever you go to store the Command, and keep the argStorage field updated before you store it to the DB.

我还是建议做第一种情况。这对以后的恋爱很有帮助。

使用一些JPA 2实现:它添加了一个@ElementCollection注释,类似于Hibernate的注释,这正是您所需要的。这里有一个例子。

Edit

正如下面评论中提到的,正确的JPA 2实现是

javax.persistence.ElementCollection

@ElementCollection
Map<Key, Value> collection;

参见:http://docs.oracle.com/javaee/6/api/javax/persistence/ElementCollection.html