java - Spring JPA REST一对多

我想通过向Person实体添加地址列表来扩展Accessing JPA Data with REST示例。因此，我添加了带有addresses批注的列表@OneToMany:

@Entity
public class Person {
    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    private long id;

    private String firstName;
    private String lastName;

    @OneToMany(fetch = FetchType.EAGER, cascade = CascadeType.ALL)
    private List<Address> addresses = new ArrayList<>();

   // get and set methods...
}

Address类是一个非常简单的类:

@Entity
public class Address {
    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    private long id;
    private String street;
    private String number;
    // get and set methods...
}

最后，我添加了AddressRepository接口(interface):

public interface AddressRepository extends PagingAndSortingRepository<Address, Long> {}

然后，我尝试发布一个带有某些地址的人:

curl -i -X POST -H "Content-Type:application/json" -d '{  "firstName" : "Frodo",  "lastName" : "Baggins", "addresses": [{"street": "somewhere", "number": 1},{"street": "anywhere", "number": 0}]}' http://localhost:8080/people

我得到的错误是:

Could not read document: Failed to convert from type [java.net.URI] to type [ws.model.Address] for value 'street';
nested exception is java.lang.IllegalArgumentException: Cannot resolve URI street. Is it local or remote? Only local URIs are resolvable. (through reference chain: ws.model.Person[\"addresses\"]->java.util.ArrayList[1]);
nested exception is com.fasterxml.jackson.databind.JsonMappingException: Failed to convert from type [java.net.URI] to type [ws.model.Address] for value 'street'; nested exception is java.lang.IllegalArgumentException: Cannot resolve URI street. Is it local or remote? Only local URIs are resolvable. (through reference chain: ws.model.Person[\"addresses\"]->java.util.ArrayList[1])

哪种方法可以创建一对多关系并将json对象发布到它们之间？

最佳答案

您应该首先发布这两个地址，然后在您的Person POST中使用返回的URL(例如http://localhost:8080/addresses/1和http://localhost:8080/addresses/2):

curl -i -X POST -H "Content-Type:application/json" -d '{  "firstName" : "Frodo",  "lastName" : "Baggins", "addresses": ["http://localhost:8080/addresses/1","http://localhost:8080/addresses/2"]}' http://localhost:8080/people

如果要先保存此人，然后添加其地址，则可以执行以下操作:

curl -i -X POST -H "Content-Type:application/json" -d '{  "firstName" : "Frodo",  "lastName" : "Baggins"}' http://localhost:8080/people
curl -i -X POST -H "Content-Type:application/json" -d '{"street": "somewhere", "number": 1}' http://localhost:8080/addresses
curl -i -X POST -H "Content-Type:application/json" -d '{"street": "anywhere", "number": 0}' http://localhost:8080/addresses
curl -i -X PATCH -H "Content-Type: text/uri-list" -d "http://localhost:8080/addresses/1
http://localhost:8080/addresses/2" http://localhost:8080/people/1/addresses

关于java - Spring JPA REST一对多，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/34583515/