如何在Spring JPA存储库中联接多个表的结果

2022-09-01 17:50:05

我是Spring的新手,我无法弄清楚如何连接多个表以返回一些结果。我试图实现一个小型的库应用程序,如下所示。

我的实体类 - 预订、客户、预订


书籍.java - 图书馆有书籍

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

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

    @NotNull(message = "Book name cannot be null")
    @Column(name = "book_name", columnDefinition = "VARCHAR(255)")
    private String bookName;

    @Column(name = "author", columnDefinition = "VARCHAR(255)")
    private String author;

    // getters and setters

    public Book() {}

    public Book(String bookName, String author) {
        this.bookName = bookName;
        this.author = author;
    }
}

客户.java - 在库中注册的客户

@Entity
@Table(name = "customer", uniqueConstraints = {@UniqueConstraint(columnNames = {"phone"})})
public class Customer {
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "id", columnDefinition = "int")
    private int id;

    @NotNull(message = "Customer name cannot be null")
    @Column(name = "name", columnDefinition = "VARCHAR(255)")
    private String name;

    @Column(name = "phone", columnDefinition = "VARCHAR(15)")
    private String phone;

    @Column(name = "registered", columnDefinition = "DATETIME")
    private String registered;

    // getters and setters

    public Customer() {}

    public Customer(String name, String phone, String registered) {
        this.name = name;
        this.phone = phone;
        this.registered = registered;
    }
}

预订.java - 客户进行的所有预订

@Entity
@Table(name = "bookings")
public class Booking {
    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    @Column(name = "id", columnDefinition = "int")
    private int id;

    @NotNull(message = "Book id cannot be null")
    @Column(name = "book_id", columnDefinition = "int")
    private int bookId;

    @NotNull(message = "Customer id cannot be null")
    @Column(name = "customer_id", columnDefinition = "int")
    private int customerId;

    @Column(name = "issue_date", columnDefinition = "DATETIME")
    private String issueDate;

    @Column(name = "return_date", columnDefinition = "DATETIME")
    private String returnDate;

    // getters and setters

    public Booking() {}

    public Booking(int bookId, int customerId, String issueDate) {
        this.bookId = bookId;
        this.customerId = customerId;
        this.issueDate = issueDate;
    }
}

现在,各个实体的表架构如下所示:

books:
+-----------+--------------+------+-----+---------+----------------+
| Field     | Type         | Null | Key | Default | Extra          |
+-----------+--------------+------+-----+---------+----------------+
| id        | int(11)      | NO   | PRI | NULL    | auto_increment |
| book_name | varchar(255) | NO   |     | NULL    |                |
| author    | varchar(255) | YES  |     | NULL    |                |
+-----------+--------------+------+-----+---------+----------------+
id - primary key

customer:
+------------+--------------+------+-----+-------------------+-------------------+
| Field      | Type         | Null | Key | Default           | Extra             |
+------------+--------------+------+-----+-------------------+-------------------+
| id         | int(11)      | NO   | PRI | NULL              | auto_increment    |
| name       | varchar(255) | NO   |     | NULL              |                   |
| registered | datetime     | YES  |     | CURRENT_TIMESTAMP | DEFAULT_GENERATED |
| phone      | varchar(15)  | YES  | UNI | NULL              |                   |
+------------+--------------+------+-----+-------------------+-------------------+
id - primary key

bookings:
+-------------+----------+------+-----+-------------------+-------------------+
| Field       | Type     | Null | Key | Default           | Extra             |
+-------------+----------+------+-----+-------------------+-------------------+
| id          | int(11)  | NO   | PRI | NULL              | auto_increment    |
| book_id     | int(11)  | NO   | MUL | NULL              |                   |
| customer_id | int(11)  | NO   | MUL | NULL              |                   |
| issue_date  | datetime | YES  |     | CURRENT_TIMESTAMP | DEFAULT_GENERATED |
| return_date | datetime | YES  |     | NULL              |                   |
+-------------+----------+------+-----+-------------------+-------------------+
id - primary key
book_id - foreign key references books.id
customer_id - foreign key references customer.id

现在我想做的是给出一些预订评论,如客户电话或作者姓名等,我想退回与该订单相关的所有预订。我将展示一个示例 Booking api 来解释。

预订控制器:

@RestController
@RequestMapping("/bookings")
public class BookingController {
    @Autowired
    BookingService bookingService;

    // some booking apis which return Booking objects

    @GetMapping
    public List<Booking> getAllBookingsBy(@RequestParam("phone") String phone,
                                         @RequestParam("authors") List<String> authors) {
        return bookingService.getAllBy(phone, authors);
    }

    @PostMapping
    public Booking addBooking(@RequestBody Booking booking) {
        bookingService.saveBooking(booking);
        return booking;
    }
}

预订服务等级:

@Service
public class BookingService {
    @Autowired
    private BookingRepository bookingRepository;

    // some booking service methods

    // get all bookings booked by a customer with matching phone number and books written by a given list of authors
    public List<Booking> getAllBy(String phone, List<String> authors) {
    return bookingRepository.queryBy(phone, authors);
    }

    public void saveBooking(Booking booking) {
        bookingRepository.save(booking);
    }
}

预订存储库类:

@Repository
public interface BookingRepository extends JpaRepository<Booking, Integer> {
    // some booking repository methods

    @Query(value = "SELECT * FROM bookings bs WHERE " +
            "EXISTS (SELECT 1 FROM customer c WHERE bs.customer_id = c.id AND c.phone = :phone) " +
            "AND EXISTS (SELECT 1 FROM books b WHERE b.id = bs.book_id AND b.author IN :authors)",
            nativeQuery = true)
    List<Booking> queryBy(@Param("phone") String phone,
                            @Param("authors") List<String> authors);
}

现在点击显示的预订控制器将返回一个如下所示的预订对象:

[
    {
        "id": 3,
        "book_id": 5,
        "customer_id": 2,
        "issue_date": "2019-02-04 01:45:21",
        "return_date": null
    }
]

但我不希望这样,我想和他们一起返回该预订的客户姓名以及书籍的名称。因此,我希望控制器返回的预订对象如下所示:

[
    {
        "id": 3,
        "book_id": 5,
        "customer_id": 2,
        "issue_date": "2019-02-04 01:45:21",
        "return_date": null,
        "customer_name": "Cust 2",
        "book_name": "Book_2_2",
    }
]

有人可以帮忙做这件事吗?我卡住了,因为我无法从这里继续。

#########编辑:我在我的预订类中添加了这些单向的一对一关联:

@OneToOne
@JoinColumn(name = "book_id", insertable = false, updatable = false)
private Book book;

@OneToOne
@JoinColumn(name = "customer_id", insertable = false, updatable = false)
private Customer customer;

但现在,当我点击控制器时,我会在 Booking 对象中获得整个 Book 和 Customer 对象。那么,我该怎么做才能在预订对象中返回书名和客户名称呢?以下是我的预订对象现在返回的样子:

[
    {
        "id": 3,
        "book_id": 5,
        "book": {
            "id": 5,
            "book_name": "Book_2_2",
            "author": "author_2"
        },
        "customer_id": 2,
        "customer": {
            "id": 2,
            "name": "Cust 2",
            "phone": "98765431",
            "registered": "2019-02-04 01:13:16"
        },
        "issue_date": "2019-02-04 01:45:21",
        "return_date": null
    }
]

此外,现在我的预订控制器中的save()api不起作用,因为当我向其发送Booking类型的对象时,bookId和customerId以某种方式显示为0,这在添加这些更改之前没有发生。


答案 1

你所做的是错误的。您正在返回Booking,并且您希望它神奇地反序列化为包含诸如“书名”之类的加入信息的实体。但在存储库上的选择查询中,您选择了预订。在您的实施过程中,Booking不保存有关该书的信息。

首先,您需要将反序列化为JSON的内容与将用作春季数据的持久性层的内容分开。

  1. 从预订到预订建立/关系作为开始。@OneToOne@OneToMany
  2. 将查询更改为对已映射为 Book 的实体/集合执行预先提取。
  3. 制作一个 POJO,并使用 JSON 注释对其进行注释,使其符合控制器返回的希望方式。
  4. 在“预订”上的持久性对象/“预订”与新创建的 POJO 之间映射

实际上,如果您映射为OneToOne,则默认初始化将变为 EAGER,因此您的查询会变得有点不必要。

如果我们假设您的映射正好位于持久层中,则查询将如下所示:

@Query(value = "SELECT * FROM bookings bs WHERE " +
            "bs.customer.phone = :phone) " +
            "AND  bs.book.author IN :authors)")

这是您来自Hibernate的映射文档>http://docs.jboss.org/hibernate/orm/5.4/userguide/html_single/Hibernate_User_Guide.html#associations


答案 2

您拥有的查询不是联接表的最佳方式。更直观的方式就是这样

SELECT * FROM bookings
WHERE customer_id in (SELECT id FROM customer WHERE phone = :phone)
 AND book_id in (SELECT id FROM books WHERE author IN :authors)

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