java - hibernate和mappedBy：是否可以自动设置外键而不设置对象之间的双向关系？

Question

欢迎，

我有 2 节课：对话和提问。一次对话有很多问题。

对话.java：

package com.jcg.jpa.mappedBy;

import java.io.Serializable;
import java.util.ArrayList;
import java.util.Collection;

import javax.persistence.CascadeType;
import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.FetchType;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.OneToMany;
import javax.persistence.Table;

@Entity
@Table(name = "CONVERSATION_TABLE")
public class Conversation implements Serializable {

private static final long serialVersionUID = 1L;

@Id
@Column(name = "CONV_ID")
@GeneratedValue(strategy = GenerationType.IDENTITY)
private int conversationId;

@Column(name = "CONV_NAME")
private String name;

@OneToMany(fetch = FetchType.EAGER, cascade = CascadeType.ALL, mappedBy="conversation")
private Collection<Question> questions = new ArrayList<Question>();

public Conversation() { }

public int getConversationId() {
    return conversationId;
}

public void setConversationId(int conversationId) {
    this.conversationId = conversationId;
}

public String getName() {
    return name;
}

public void setName(String name) {
    this.name = name;
}

public Collection<Question> getQuestions() {
    return questions;
}

public void setQuestions(Collection<Question> questions) {
    this.questions = questions;
}

@Override
public String toString() {
    return "Employee [conversationId=" + conversationId + ", name=" + name + "]";
}
}

问题.java：

package com.jcg.jpa.mappedBy;

import java.io.Serializable;

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.ManyToOne;
import javax.persistence.Table;

@Entity
@Table(name = "QUESTION_TABLE")
public class Question implements Serializable {

private static final long serialVersionUID = 1L;

@Id
@Column(name = "id")
@GeneratedValue(strategy = GenerationType.IDENTITY)
private int id;

@ManyToOne
@JoinColumn(name = "CONVERSATION_CONV_ID", nullable = false)
private Conversation conversation;


@Column(name = "QUESTION_TEXT")
private String questionText;

@Column(name = "ANSWER_TEXT")
private String answerText;



public Question() { }

public int getId() {
    return id;
}

public void setId(int id) {
    this.id = id;
}

public String getQuestionText() {
    return questionText;
}

public void setQuestionText(String questionText) {
    this.questionText = questionText;
}

public String getAnswerText() {
    return answerText;
}

public void setAnswerText(String answerText) {
    this.answerText = answerText;
}

public Conversation getConversation() {
    return conversation;
}

public void setConversation(Conversation conversation) {
    this.conversation = conversation;
}

@Override
public String toString() {
    return "Question [id=" + id +  ", questionText=" + questionText
            + ", answerText=" + answerText +"]";
}
}

持久性.xml：

<?xml version="1.0" encoding="UTF-8"?>
<persistence version="2.1"
xmlns="http://xmlns.jcp.org/xml/ns/persistence" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://xmlns.jcp.org/xml/ns/persistence http://xmlns.jcp.org/xml/ns/persistence/persistence_2_1.xsd">
<persistence-unit name="JPAMappedbyExample" transaction-type="RESOURCE_LOCAL">
    <class>com.jcg.jpa.mappedBy.Conversation</class>
    <class>com.jcg.jpa.mappedBy.Question</class>

    <!-- Configuring The Database Connection Details -->
    <properties>
        <property name="javax.persistence.jdbc.driver" value="com.mysql.jdbc.Driver" />
        <property name="javax.persistence.jdbc.url" value="jdbc:mysql://localhost:3306/jpatest" />
        <property name="javax.persistence.jdbc.user" value="root" />
        <property name="javax.persistence.jdbc.password" value="qwerty" />

    </properties>
</persistence-unit>

现在在 Main.java 我试图用两个问题创建对话：

package com.jcg.jpa.mappedBy;

import java.util.ArrayList;
import java.util.List;

import javax.persistence.EntityManager;
import javax.persistence.EntityManagerFactory;
import javax.persistence.Persistence;

public class Main {

private static final EntityManagerFactory emFactoryObj;
private static final String PERSISTENCE_UNIT_NAME = "JPAMappedbyExample";   

static {
    emFactoryObj = Persistence.createEntityManagerFactory(PERSISTENCE_UNIT_NAME);
}

// This Method Is Used To Retrieve The 'EntityManager' Object
public static EntityManager getEntityManager() {
    return emFactoryObj.createEntityManager();
}

private static void insertRecords() {
    EntityManager entityMgrObj = getEntityManager();
    if (null != entityMgrObj) {
        entityMgrObj.getTransaction().begin();

        Conversation conv = new Conversation();
        conv.setName("Discussion about something");

        Question question1 = new Question();
        question1.setQuestionText("2 plus 2");
        question1.setAnswerText("four");
        question1.setConversation(conv);

        Question question2 = new Question();
        question2.setQuestionText("what is Your name");
        question2.setAnswerText("Adam");
        question2.setConversation(conv);

        List<Question> questions = new ArrayList<Question>();
        questions.add(question1);
        questions.add(question2);
        conv.setQuestions(questions);

        entityMgrObj.persist(conv);
        entityMgrObj.getTransaction().commit();

        entityMgrObj.clear();
        System.out.println("Record Successfully Inserted In The Database");
    }
}



public static void main(String[] args) {
    insertRecords();

}
}

在 insertRecords() 我正在创建 conv 和两个问题。每个问题都设置了对话：

question1.setConversation(conv);
question2.setConversation(conv);

接下来，创建一个包含这 2 个问题的问题列表并将其设置为 conv questions list：

conv.setQuestions(questions);

它工作正常，因为数据被插入到两个表中，并且外键 CONVERSATION_CONV_ID 被填充：

但是，当我删除问题中的设置对话时，行：

question1.setConversation(conv);
question2.setConversation(conv);

外键设置为 NULL。为什么？我们已经在对话问题列表中添加了两个问题：

@OneToMany(fetch = FetchType.EAGER, cascade = CascadeType.ALL, mappedBy="conversation")
private Collection<Question> questions = new ArrayList<Question>();

，所以hibernate应该知道这两个问题的会话外键是什么（因为它们位于指定的会话问题列表中）。那么是否可以避免为每个问题设置 Conversation 并且只将问题添加到 Conversation 的问题列表中？我应该如何配置实体来做到这一点？或者也许这是不可能的，我们总是需要设置它的两个方向？

score 1 · Accepted Answer

所以hibernate应该知道这两个问题的会话外键是什么（因为它们位于指定的会话问题列表中）。

如果您不使用指定Hibernate，则任何 Hibernate 都不应该知道Conversation这两个是什么，因为它在这里处理对象，而这两个没有任何迹象表明它们属于哪个部分。QuestionssetConversation()questionsConversation

解释：

因为当您只将这两个添加questions到Conversation对象时，此信息将不会在Questions对象中可见。

这里的另一件事mappedBy是，该属性指示object映射的所有者，那么如果侧面没有给定对象，如何进行映射mappedBy？

这就是您应该指定ConversationinQuestion对象的原因，以便可以正确评估映射Hibernate。

笔记：

建议使用SetforOneToMany映射而不是任何其他Collection的，因为这个集合不应该有任何重复，所以你最好把它改成：

@OneToMany(fetch = FetchType.EAGER, cascade = CascadeType.ALL, mappedBy="conversation")
private Set<Question> questions = new HashSet<Question>();

java - hibernate和mappedBy：是否可以自动设置外键而不设置对象之间的双向关系？

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