@OneToOne 关系 Spring 数据 JPA/Hibernate 实体关系

Question

I'm working on adding a feature to an already developed spring boot web application. The primary entity that has child entities is a Record. It has a few columns/variables that I want to now be in its own, separate entity (CustomerOrder) and exist in a one-to-one relationship with the Record. To summarize:

Record {

• thing 1
• thing 2
• thing 3

}

is now becoming:

CustomerOrder {

• thing 1
• thing 2
• thing 3

}

Record { CustomerOrder }

I'm having some issues with what I've produced. Here is the CustomerOrder model's relevant relationship data:

@Entity
@Table(name="customer_orders")
public class CustomerOrder {

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

    ... other columns

    @OneToOne(orphanRemoval = true, cascade = CascadeType.ALL, mappedBy="customerOrder", fetch = FetchType.EAGER)
    private Record record;


}

And then here is the Record model's relevant data:

@Entity
@Table(name="records")
public class Record extends Auditable<String> implements Serializable {

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

    ... other columns

    @OneToOne
    @JoinColumn(name="customer_order_id", nullable = false, unique = true)
    private CustomerOrder customerOrder;
}

My issue exists when I try to POST a record, when a user tries creating one in the ui. Here is the POST method for a record:

    @PostMapping
    public ResponseEntity<?> saveRecord(@RequestBody Record recordBody, BindingResult result) {
        if(!result.hasErrors()) {
            if(recordBody.getHardwareItems().isEmpty()) {
                record = recordsService.save(recordBody);
            } else {
                // Save the record first, recordId is required on hardwareItems
                // TODO: investigate Spring Hibernate/JPA rules - is there a way to save parent before children to avoid a null recordId
                CustomerOrder customerOrder = recordBody.getCustomerOrder();
                recordBody.setCustomerOrder(new CustomerOrder());
                customerOrder.setRecord(record);
                customerOrder = customerOrdersService.save(customerOrder);
                record = recordsService.save(recordBody);
            }
        } else {
            return new ResponseEntity<>(result.getAllErrors(), HttpStatus.BAD_REQUEST);
        }
        // Return the location of the created resource
        uri = ServletUriComponentsBuilder.fromCurrentRequest().path("/{recordId}").buildAndExpand(record.getId()).toUri();
        return new ResponseEntity<>(uri, HttpStatus.CREATED);
    }

The error I receive is the following:

2021-02-19 00:46:28.398  WARN 29990 --- [io-8080-exec-10] o.h.engine.jdbc.spi.SqlExceptionHelper   : SQL Error: 1364, SQLState: HY000
2021-02-19 00:46:28.398 ERROR 29990 --- [io-8080-exec-10] o.h.engine.jdbc.spi.SqlExceptionHelper   : Field 'record_id' doesn't have a default value

This makes sense to me at least, since I'm trying to save the CustomerOrder object that depends on a Record object, which has yet to have been persisted. So, how do I go about changing up the order and/or creating and persisting a Record object so that I can then save the CustomerOrder object to it?

Answer 1

You need to mark your column record_id as AI(AUTO_INCREMENT) in your table definition.

ALTER TABLE records CHANGE record_id INT(6) NOT NULL AUTO_INCREMENT;

Your primary key is record_id , add @Column(name = "record_id", nullable = false)

@Entity
@Table(name="records")
public class Record extends Auditable<String> implements Serializable {

    @Id
    @GeneratedValue(strategy=GenerationType.IDENTITY)
    @Column(name = "record_id", nullable = false)
    private Long id;

    ... other columns

    @OneToOne
    @JoinColumn(name="customer_order_id", nullable = false, unique = true)
    private CustomerOrder customerOrder;
}