休眠持續單向一對多關系

Question

我有以下模型：

對應的映射如下：

報價實體

@Entity
@Table(name="Quote")
@Getter
@Setter
public class Quote {

    @Id
    @GeneratedValue(strategy= GenerationType.AUTO, generator="native")
    @GenericGenerator(name = "native", strategy = "native")
    @Column(name="idQuote")
    private int id;

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

    @OneToMany(cascade = CascadeType.ALL, orphanRemoval = true)
    @JoinColumn(name="Quote_idQuote")
    private Set<Item> item;
}

項目實體

@Entity
@Table(name="Item")
@Getter
@Setter
public class Item {

    @Id
    @GeneratedValue(strategy= GenerationType.AUTO, generator="native")
    @GenericGenerator(name = "native", strategy = "native")
    @Column(name="idItem")
    private int id;

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

    @Column(name="Quote_idQuote")
    private int quoteId;
}

我的問題是我無法使用項目列表來保留Quote，這是我嘗試執行的代碼：

SessionFactory sessionFactory;
Configuration configuration= new Configuration();
configuration.configure();
sessionFactory=configuration.buildSessionFactory();
Session session= sessionFactory.openSession();
session.beginTransaction();

Quote quote= new Quote();
quote.setNumber("ASR3E4E");

Set<Item> items= new HashSet<Item>();

Item item1= new Item();
item1.setName("Item 1");

Item item2= new Item();
item1.setName("Item 2");

items.add(item1);
items.add(item2);        

quote.setItem(items);

session.persist(quote);

session.getTransaction().commit();

我收到以下錯誤：

Hibernate: insert into Quote (number) values (?)
Hibernate: insert into Item (name, Quote_idQuote) values (?, ?)
Aug 09, 2018 3:27:25 PM org.hibernate.engine.jdbc.spi.SqlExceptionHelper logExceptions
WARN: SQL Error: 1452, SQLState: 23000
Aug 09, 2018 3:27:25 PM org.hibernate.engine.jdbc.spi.SqlExceptionHelper logExceptions
ERROR: Cannot add or update a child row: a foreign key constraint fails (`test`.`item`, CONSTRAINT `fk_Item_Quote` FOREIGN KEY (`Quote_idQuote`) REFERENCES `quote` (`idquote`))
Exception in thread "main" javax.persistence.PersistenceException: org.hibernate.exception.ConstraintViolationException: could not execute statement
    at org.hibernate.internal.ExceptionConverterImpl.convert(ExceptionConverterImpl.java:149)
    at org.hibernate.internal.ExceptionConverterImpl.convert(ExceptionConverterImpl.java:157)
    at org.hibernate.internal.ExceptionConverterImpl.convert(ExceptionConverterImpl.java:164)
    at org.hibernate.internal.SessionImpl.firePersist(SessionImpl.java:831)

鑒於以上所述，這讓我開始思考...如何做到這一點，以便自動休眠可以將Quote及其項目持久化。 為什么不休眠首先生成Quote的ID，將其保留，然后再保留Items？ 我應該如何進行映射以保持單向關系，並且我可以毫無問題地完成我想做的事情？

非常感謝！

Answer 1

映射問題中顯示的數據庫模型的正確方法如下：

報價實體

@Entity
@Table(name="Quote")
@Getter
@Setter
public class Quote {

    @Id
    @GeneratedValue(strategy= GenerationType.AUTO, generator="native")
    @GenericGenerator(name = "native", strategy = "native")
    @Column(name="idQuote")
    private int id;

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

    @OneToMany(cascade = CascadeType.ALL, orphanRemoval = true)
    @JoinColumn(name = "Quote_idQuote", nullable=false)
    private Set<Item> item;
}

項目實體

@Entity
@Table(name="Item")
@Getter
@Setter
public class Item {

    @Id
    @GeneratedValue(strategy= GenerationType.AUTO, generator="native")
    @GenericGenerator(name = "native", strategy = "native")
    @Column(name="idItem")
    private int id;

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

}

以及用於使用Quotes持久引用實體的代碼：

SessionFactory sessionFactory;
Configuration configuration= new Configuration();
configuration.configure();
sessionFactory=configuration.buildSessionFactory();
Session session= sessionFactory.openSession();
session.beginTransaction();

Quote quote= new Quote();
quote.setNumber("ASR3E4E");

Set<Item> items= new HashSet<Item>();

Item item1= new Item();
item1.setName("Item 1");

Item item2= new Item();
item2.setName("Item 2");

items.add(item1);
items.add(item2);        

quote.setItem(items);

session.persist(quote);

session.getTransaction().commit();

Answer 2

Hibernate提供了雙方映射，以實現靈活的代碼。 如果要從有價值的實體（項目）執行任何操作。 做正確的映射

1）Quote.java類

    @OneToMany(cascade = CascadeType.ALL, mappedBy="quote",fetch=FetchType.Eager)

private Set<Item> item;

2）在Item.java類中

    @ManyToOne
    @JoinColumn(name="Quote_idQuote") 
    private Quote quote;  

Answer 3

嘗試

   item1.setQuote(quote);
   item2.setQuote(quote);
   session.persist(quote);

在您的情況下，無法成功，因為您無法在此處設置關系：如果滿足以下條件，它將起作用（並且仍然非常糟糕）：

刪除級聯= CascadeType.ALL，

session.persist(quote); // this must be done first to actually get bloody quote ID that you can propagate
item1.setQuoteId(quote.getId()); //or something to set quoteid
item2.setQuoteId(quote.getId())=quote.id; //or something to set quoteid
session.persist(item1);
session.persist(item2);