为什么在使用@ManyToOne时不能在数据库中插入行？

如何解决为什么在使用@ManyToOne时不能在数据库中插入行？

我有这些实体。

@Entity
@Data
@AllArgsConstructor
@NoArgsConstructor
public class Language {
    
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Long id;
    
    @NotNull
    private String language;
    
}

和

@Entity
@Data
@AllArgsConstructor
@NoArgsConstructor
public class Sentence {
    
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Long id;
    
    @NotNull
    private String wordInfrench;
    
    @NotNull
    private String wordInOtherLanguage;
    
    @ManyToOne
    @NotNull
    private Language language;
    
}

当我尝试通过使用以下代码将行插入sentence数据库中时：

// My repository
private SentenceRepository sentenceRepository;

// Object information
Long id = 0L; 
String wordInfrench = "Oui oui"
String wordInOtherLanguage = "Ja ja"
Language language = new Language(id,"Swedish");

// Save to database
sentenceRepository.save(new Sentence(id,wordInfrench,wordInOtherLanguage,language));

我收到此错误：

org.springframework.orm.jpa.JpaObjectRetrievalFailureException: Unable to find 
    se.danielmartensson.entity.Language with id 0; nested exception is javax.persistence.EntityNotFoundException: Unable to find se.danielmartensson.entity.Language with id 0

    Caused by: javax.persistence.EntityNotFoundException: Unable to find se.danielmartensson.entity.Language with id 0

如果我使用Long id = 1L;，也会出现此错误：

org.springframework.orm.jpa.JpaObjectRetrievalFailureException: Unable to find 
se.danielmartensson.entity.Language with id 1; nested exception is javax.persistence.EntityNotFoundException: Unable to find se.danielmartensson.entity.Language with id 1

Caused by: javax.persistence.EntityNotFoundException: Unable to find se.danielmartensson.entity.Language with id 1

那是怎么回事？为什么不能将sentence插入数据库，而language数据库也将更新？

解决方法

问题是Hibernate试图在数据库中找到id = 0（和1）的Language实例，但是它不存在。

首先，您必须插入语言，然后才能插入句子：

// My repositories
private LanguageRepository languageRepository;
private SentenceRepository sentenceRepository;

// Inserting the Language
Language language = new Language();
language.setLanguage("Swedish");
// Here,note that I'm not forcing the value of the attribute "id". 
// The "@GeneratedValue" will generate one for me.

// Also note that the "save" method returns an object. In this case,// it will be a "Language" instance with the new id
language = languageRepository.save(language);

// Inserting the Sentence
Sentence sentence = new Sentence();
sentence.setWordInFrench("Oui oui");
sentence.setWordInOtherLanguage("Ja ja");
sentence.setLanguage(language);

sentenceRepository.save(sentence);

或者，您可以执行以下操作：

@Entity
@Data
@AllArgsConstructor
@NoArgsConstructor
public class Sentence {
    
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Long id;
    
    @NotNull
    private String wordInFrench;
    
    @NotNull
    private String wordInOtherLanguage;
    
    @ManyToOne(cascade = CascadeType.PERSIST)
    @JoinColumn(name = "id_language",nullable = false)
    private Language language;

我们使用cascade = CascadeType.PERSIST表示每次尝试插入新句子时，Hibernate都会自动检查数据库中是否已经存在language对象。否则，将language预先插入。此外，对于外键，建议使用@JoinColumn映射来配置列定义，例如“名称”，“可空”，“唯一”等。