加入与@ManyToMany协会三个表

Question

加入三个表与@ManyToMany协会

我实现使用Spring引导和Spring数据JPA书店应用。我曾与@OneToOne和@ManyToOne和@OneToMany协会试过，但我坚持用@ManyToMany关联。

有三个表 - 书，流派和作者。我要做到以下几点1）创建一个新表 - Book_Genre_Author 2）如果作者或类型的数据库已经存在，它不会使数据库中的一个新条目。 3）如果本书在数据库中已经存在，它不会使一个进入数据库。 4）如何使用Spring数据JPA实现插入和更新操作？

这里是我的三个实体 -

@Entity
@Table(name = "Book_Details")
public class Book {

@Id
@GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "CUSTOM_SEQUENCE")
@SequenceGenerator(sequenceName = "customer_seq", allocationSize = 1, name = "CUSTOM_SEQUENCE")
@Column(name = "BookId")
int bookId;

public int getBookId() {
    return bookId;
}

public void setBookId(int bookId) {
    this.bookId = bookId;
}

@Column(name = "BookName")
String bookName;



@ManyToMany(cascade=CascadeType.ALL)
@JoinTable(name = "Book_Author_Genre", joinColumns = @JoinColumn(name = "bookId"), inverseJoinColumns = @JoinColumn(name = "genreId"))
@MapKeyJoinColumn(name = "authorId")
@ElementCollection
private Map<Author, Genre> authorGenre = new HashMap<>();

public Map<Author, Genre> getAuthorGenre() {
    return authorGenre;
}

public void setAuthorGenre(Map<Author, Genre> authorGenre) {
    this.authorGenre = authorGenre;
}

@Column(name = "Language")
String language;



@Column(name = "GoodReadReviews")
String goodReadReviews;


  @Override public String toString() { return "BookId=" + this.bookId +
  "::BookName=" + this.bookName + "::AuthorId=" + "Will check how to get it" +
  "::Langauge = " + this.language + "::GenreId = " + "Will check how to get it" +
  "::GoodReadReviews=" + this.goodReadReviews;

  }


public String getBookName() {
    return bookName;
}

public void setBookName(String bookName) {
    this.bookName = bookName;
}

public String getLanguage() {
    return language;
}

public void setLanguage(String language) {
    this.language = language;
}

public String getGoodReadReviews() {
    return goodReadReviews;
}

public void setGoodReadReviews(String goodReadReviews) {
    this.goodReadReviews = goodReadReviews;
}

}

@Entity
@Table(name="Author_Details")
public class Author {

@Id
@GeneratedValue(strategy = GenerationType.SEQUENCE, generator="AuthorSeqGenerator")
@SequenceGenerator(sequenceName = "author_seq", allocationSize = 1, name = "AuthorSeqGenerator")
@Column(name="AuthorId")
int authorId;

@Column(name="AuthorName")
String authorName;

@Column(name="Country")
String country;

@Column(name="Gender")
String gender;


public int getAuthorId() {
    return authorId;
}
public void setAuthorId(int authorId) {
    this.authorId = authorId;
}


public String getAuthorName() {
    return authorName;
}
public void setAuthorName(String authorName) {
    this.authorName = authorName;
}
public String getCountry() {
    return country;
}
public void setCountry(String country) {
    this.country = country;
}
public String getGender() {
    return gender;
}
public void setGender(String gender) {
    this.gender = gender;
}

}

@Entity
@Table(name="Genre_details")
public class Genre {

@Id
@GeneratedValue(strategy = GenerationType.SEQUENCE, generator="GenreSeqGenerator")
@SequenceGenerator(sequenceName = "genre_seq", allocationSize = 1, name = "GenreSeqGenerator")
int genreId;

@Column(name="GenreName")
String genreName;

public int getGenreId() {
    return genreId;
}

public void setGenreId(int genreId) {
    this.genreId = genreId;
}




public String getGenreName() {
    return genreName;
}

public void setGenreName(String genreName) {
    this.genreName = genreName;
}

}

我试着用下面的代码保存书籍的详细信息 -

Author author = new Author();
        author.setAuthorName("Rabindranath Tagore");
        author.setCountry("India");
        author.setGender("Male");

    Genre genre = new Genre();
    genre.setGenreName("Fiction");

    Book book = new Book();
    book.setBookName("Gitanjali");
    book.setGoodReadReviews("5");
    book.setLanguage("Bengali");
    HashMap<Author,Genre> authorGenre = new HashMap<>();
    authorGenre.put(author, genre);
    book.setAuthorGenre(authorGenre);

    bookRepository.save(book);

不过，我收到以下错误。

org.springframework.dao.InvalidDataAccessApiUsageException：由造成org.hibernate.TransientObjectException：对象引用一个未保存的瞬态的实例 - 冲洗之前保存的瞬态的实例：com.bsm.app.model.Author;嵌套的例外是java.lang.IllegalStateException：org.hibernate.TransientObjectException：对象引用一个未保存的瞬态的实例 - 冲洗之前保存的瞬态的实例：com.bsm.app.model.Author

如我在它天真的我可能是错误的，我的做法。请建议和支持。

Answer 1

我建议建立这将被用于映射特定表的新链接的实体。

你只需要决定是否要留在一个复合主键或添加人工ID（推荐）。

这种设置似乎涉及更多的代码，但是从我的经验更为清晰，一旦3个表开始发挥作用：

class Book {
    @OneToMany(mappedBy = "book", cascade = {CascadeType.PERSIST, CascadeType.MERGE})
    private Set<BookAuthorGenreLink> bookAuthorGenreLinks;
}

class Author{
    @OneToMany(mappedBy = "author", cascade = {CascadeType.PERSIST, CascadeType.MERGE})
    private Set<BookAuthorGenreLink> bookAuthorGenreLinks;
}

class Genre{
    @OneToMany(mappedBy = "genre", cascade = {CascadeType.PERSIST, CascadeType.MERGE})
    private Set<BookAuthorGenreLink> bookAuthorGenreLinks;
}

链接表：

class BookAuthorGenreLink {

    @Id
    @GeneratedValue
    private Long id;

    @ManyToOne(cascade = {CascadeType.PERSIST, CascadeType.MERGE})
    @JoinColumn(name = "book_id")
    private Book book;

    @ManyToOne(cascade = {CascadeType.PERSIST, CascadeType.MERGE})
    @JoinColumn(name = "author_id")
    private Author author;

    @ManyToOne(cascade = {CascadeType.PERSIST, CascadeType.MERGE})
    @JoinColumn(name = "genre_id")
    private Genre genre;

}