java—如何在SpringJPA存储库中连接多个表的结果

41zrol4v  于 2021-07-09  发布在  Java
关注(0)|答案(3)|浏览(332)

我是spring新手,我不知道如何连接多个表以返回一些结果。我尝试实现一个小的库应用程序,如下所示。
我的实体类-book、customer、bookings
book.java-图书馆提供的书籍

@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;
    }
}

customer.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;
    }
}

booking.java-客户的所有预订

@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,各个实体的表模式如下所示:

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

现在我想做的是给一些预订的关键,如客户电话或作者姓名等,我想返回所有的预订相关的订单。我将展示一个示例预订api来解释。
售票员:

@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booking controller将返回如下所示的booking对象:

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

但我不想这样,我想和他们一起把订票的顾客的名字和书的名字还给他们。所以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",
    }
]

有人能帮忙吗?我被困住了,因为我无法从这里开始。

编辑:我在预订课程中添加了这些单向onetoone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,我的预订对象中包含了整本书和客户对象。那么我该怎么做才能在预订对象中返回bookname和customer name呢?下面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
    }
]

另外,我的booking controller中的save()api也不起作用,因为当我向它发送booking类型的对象时,bookid和customerid以某种方式显示为0,这在我添加这些更改之前是没有发生的。

yhxst69z

yhxst69z1#

您可以按照以下步骤来实现。
为响应所需的所有字段创建一个带有getter的新接口。
在@query内的查询字符串中,需要为select中的列提供名称。注意:这些名称需要与您在接口中创建的getter同步。
将此接口用作存储库方法的返回类型。
有关更多信息,可以参考spring data rest中的投影。https://docs.spring.io/spring-data/jpa/docs/current/reference/html/#projections

5m1hhzi4

5m1hhzi42#

您拥有的查询不是联接表的最佳方式。更直观的方法是这样的

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)
b4lqfgs4

b4lqfgs43#

你所做的是错误的。您正在返回booking,并希望它magicaly反序列化为一个实体,该实体包含图书名称等连接信息。但是在您对存储库的select查询中,您选择了预订。按照你的执行方式,预订并不包含关于这本书的信息。
首先,您需要将反序列化为json的内容和用作spring数据持久层的内容分开。
制造 @OneToOne / @OneToMany 从预订到预订的关系开始。
将查询更改为对已Map为book的实体/集合执行快速获取。
制作一个pojo并用json注解对其进行注解,您希望控制器返回它。
在您的持久化对象/预订与hidrated collection on book和新创建的pojo之间进行Map
实际上,如果您Map为onetoone,那么默认的初始化就变得非常迫切,因此您的查询就变得有点不必要了。
如果我们假设您的Map正好位于持久层,那么您的查询将如下所示:

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

这是hibernate的Map文档>http://docs.jboss.org/hibernate/orm/5.4/userguide/html_single/hibernate_user_guide.html#associations

相关问题