JoinTable没有一个常规的L​​ong id，而是EmbeddedId，我该如何创建一个Spring存储库？ [英] JoinTable doesn't have a regular Long id, but rather EmbeddedId instead, how do I create a Spring repository with that?

问题描述

根据@Vlad Mihalcea提供的示例，我已经实现了@ManyToMany与EmbeddedId类的映射。它可以工作，但是我需要为这个JoinTable实体创建和使用Spring的REST存储库。

通常，我会这样做来启用存储库：

  @CrossOrigin 
 @RepositoryRestResource（collectionResourceRel =companyproducts，path =companyproducts）
 public interface CompanyProductRepository 
扩展了PagingAndSortingRepository< CompanyProduct，CompanyProduct.CompanyProductId> {
 
} 
  

但是，由于在此实体（JoinTable）没有经典的 id ，就像这样：

  public class FooBar {
 .... 
 
 @Id 
 @GeneratedValue（strategy = GenerationType.IDENTITY）
 private长ID; 
 
 ... 
} 
  

但是我有：

  public class FooBar {
 .... 
 
 @EmbeddedId 
私人FooBarId ID; 
 
 ... 
} 
  

我无法使用 Long 或者让我们说 int ，因为标识符不是这些类型中的任何一个，它基本上是类型 FooBarId 。

顺便说一下，我还会提供大部分的类：

< pre $ @Embeddable
public class FooBarId implements Serializable {

@Column（name =foo_id）
private Long fooId;

@Column（name =bar_id）
private长barId;

private FooBarId（）{}

public FooBarId（Long fooId，Long barId）{
this.fooId = fooId;
this.barId = barId;
}
...
}


当然，在存储库声明中放置 FooBarId 而不是 Long ，但是在我的REST json响应中，这：

  http：// localhost：8080/api/foobar/com.example.springrest.entities.foobarid@g43 
  

（注意类本身的路​​径，而不是数字参数，例如5）。

我的问题是：是否可以使用 EmbeddedId 而不是 Long ， int 等等。

解决方案

当然

  public interface FooBarRepository extends PagingAndSortingRepository< FooBar，FooBarId> {} 
  

id是可嵌入类，因此可以使用它来代替经典基元。

响应中的问题可能与您的保存/检索逻辑有关

编辑：基于响应是类，而不是数字我期望你从存储库中检索到的id，但你显然应该从id获得数值，因为现在这个id是可嵌入类

I've implemented @ManyToMany mapping with an EmbeddedId class, as per the example provided by @Vlad Mihalcea. It works, however I need to create and use Spring's REST repository for this JoinTable Entity.

Usually, I would do something like this to enable the repository:

@CrossOrigin
@RepositoryRestResource(collectionResourceRel = "companyproducts", path = "companyproducts")
public interface CompanyProductRepository
    extends PagingAndSortingRepository<CompanyProduct, CompanyProduct.CompanyProductId> {

}

But, since in this Entity (JoinTable) I don't have a classic id, like this:

public class FooBar {
    ....

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Long id;    

    ...
}

but instead I have:

public class FooBar {
    ....

    @EmbeddedId
    private FooBarId id;    

    ...
}

I cannot use Long or let's say int because the identifier is not of any of those types, it's basically of type FooBarId.

By the way, I will also provide most of that class:

@Embeddable
public class FooBarId implements Serializable {

    @Column(name = "foo_id")
    private Long fooId;

    @Column(name = "bar_id")
    private Long barId;

    private FooBarId() {}

    public FooBarId(Long fooId, Long barId) {
        this.fooId= fooId;
        this.barId= barId;
    }
...
}

I did try, of course, putting FooBarId instead of Long in the repository declaration, but then in my REST json response I get something like this:

http://localhost:8080/api/foobar/com.example.springrest.entities.foobarid@g43

(note the path to the class itself, not a numeral parameter, eg. 5).

My question is: Is it possible to use repositories with an EmbeddedId instead of Long, int, etc.?

解决方案

Yes of course

public interface FooBarRepository extends PagingAndSortingRepository<FooBar, FooBarId > { }

the id is the embeddable class so use that instead of the classic primitive.

the problem in the response is probably something related to your save/retrieve logic

Edit: based on the response being the class instead of the numeral i expect you retrieved the id from the repository, but you should obviously get the numerical value from the id since now that id is the embeddable class

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