Spring Data JPA不持久化oneToMany列表

Question

我正在创建一个新项目，并使用Spring Data JPA创建一些RESTendpoint。

<parent>
<groupId>org.springframework.boot</groupId>
<artifactId>spring-boot-starter-parent</artifactId>
<version>2.1.6.RELEASE</version>
<relativePath/> <!-- lookup parent from repository -->
</parent>

只要json文件没有任何oneToMany数据，我就可以将其放到并持久化到我的主类（customer）中。然而，当张贴给客户，如果有任何数据，我会得到错误。

@Entity
@Table(name="customer")
@EntityListeners(AuditingEntityListener.class)
public class Customer extends Auditable<String> {

@Id
@GeneratedValue(strategy=GenerationType.IDENTITY)
@Column(name="id")
private int id;
@Column(name="first_name")
private String firstName;
@Column(name="last_name")
private String lastName;

@OneToMany(fetch=FetchType.LAZY, mappedBy="customer", cascade={CascadeType.ALL})
private List<EmailAddress> emailAddresses;
.......

那些电子邮件

@Table(name="email_address")
@EntityListeners(AuditingEntityListener.class)
public class EmailAddress extends Auditable<String> {

@Id
@GeneratedValue(strategy=GenerationType.IDENTITY)
@Column(name="id")
private int id;
@Column(name="email_type")
private byte emailType;
@Column(name="email")
private String email;

@ManyToOne(fetch=FetchType.LAZY, cascade={CascadeType.ALL})
@JoinColumn(name="customer_id")
@JsonIgnore
private Customer customer;
.....

postman json测试

{
    "id": 1,
    "firstName": "Bobby",
    "lastName": "Smith",
    "emailAddresses": [
        {
            "id": 1,
            "emailType": 1,
            "email": "[e-mail]",
        },
        {
            "id": 2,
            "emailType": 1,
            "email": "[e-mail]",
        }
    ]

}

顺便说一句，我已经确认在客户控制器中，电子邮件包括在客户的请求体中。

客户控制器

@PutMapping("/customers")
public Customer updateCustomer(@RequestBody Customer theCustomer) {

System.out.println("****email count "+theCustomer.getEmailAddresses().size());
for(EmailAddress index: theCustomer.getEmailAddresses()) {
System.out.println(index.toString());
}

customerService.save(theCustomer);

return theCustomer;
}

@Override
public void save(Customer theCustomer) {
//Validate the input
if(theCustomer == null) {
throw new CustomerNotFoundException("Did not find the Customer, was null...");
}
customerRepository.save(theCustomer);
}

--
-- Table structure for table `customer`
--
DROP TABLE IF EXISTS `customer`;
CREATE TABLE `customer` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`first_name` varchar(24) COLLATE utf8_bin NOT NULL,
`last_name` varchar(24) COLLATE utf8_bin NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=6 DEFAULT CHARSET=utf8 COLLATE=utf8_bin COMMENT='Primary Customer Table';

--
-- Table structure for table `email_address`
--
DROP TABLE IF EXISTS `email_address`;
CREATE TABLE `email_address` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `email_type` tinyint(4) unsigned NOT NULL COMMENT 'email type',
  `email` varchar(128) COLLATE utf8_bin NOT NULL COMMENT 'email address',
  `customer_id` int(11) NOT NULL COMMENT 'foreign key',
  INDEX par_ind (customer_id),
  PRIMARY KEY (`id`),
  UNIQUE KEY `email` (`email`),
  KEY FK_EMAIL_CUSTOMER_idx (customer_id),
  CONSTRAINT FK_EMAIL_CUSTOMER FOREIGN KEY (customer_id) REFERENCES customer (id) ON DELETE CASCADE ON UPDATE CASCADE
) ENGINE=InnoDB AUTO_INCREMENT=9 DEFAULT CHARSET=utf8 COLLATE=utf8_bin COMMENT='email addresses';

{
    "status": 400,
    "message": "could not execute statement; SQL [n/a]; constraint [null]; nested exception is org.hibernate.exception.ConstraintViolationException: could not execute statement",
    "timeStamp": 1566840491483
}

****email count 2
EmailAddress [id=1, type=1, email=[e-mail]]
EmailAddress [id=2, type=1, email=[e-mail]]
2019-08-28 17:33:07.625  WARN 8669 --- [nio-8080-exec-2] o.h.engine.jdbc.spi.SqlExceptionHelper   : SQL Error: 1048, SQLState: 23000
2019-08-28 17:33:07.626 ERROR 8669 --- [nio-8080-exec-2] o.h.engine.jdbc.spi.SqlExceptionHelper   : Column 'customer_id' cannot be null
2019-08-28 17:33:07.629 ERROR 8669 --- [nio-8080-exec-2] o.h.i.ExceptionMapperStandardImpl        : HHH000346: Error during managed flush [org.hibernate.exception.ConstraintViolationException: could not execute statement]
2019-08-28 17:33:07.735  WARN 8669 --- [nio-8080-exec-2] .m.m.a.ExceptionHandlerExceptionResolver : Resolved [org.springframework.dao.DataIntegrityViolationException: could not execute statement; SQL [n/a]; constraint [null]; nested exception is org.hibernate.exception.ConstraintViolationException: could not execute statement]

因此，对于post或put，我不确定为什么Spring Data JPA save不能满足具有oneToMany关系的实体的外键约束。我猜这可能是一些缺少的注释，或者是我的sql脚本出了问题。不确定为什么更新数据没有持久化到email_address表。emailAddress实体是否需要某种类型的customer_id getter/setter？

Answer 1

public class Customer extends Auditable<String> {


    @OneToMany(fetch=FetchType.LAZY, mappedBy="customer", cascade={CascadeType.ALL})
    private List<EmailAddress> emailAddresses;

} 

public class EmailAddress extends Auditable<String> {

   @ManyToOne(fetch=FetchType.LAZY, cascade={CascadeType.ALL})
   @JoinColumn(name="customer_id")

   private Customer customer;
}

这里的mappedby意味着Customer和EmailAddress之间的关系（即Custome表中Customer_ID的值）是由emailadress#cutomer确定的，而不是由Custome#emailadress确定的。

您试图显示的内容只是customer#emailaddress的内容，Hibernate在决定为该关系更新/插入哪些DB值时会忽略这些内容。因此，必须确保emailaddress#customer设置正确。

例如，您可以使用以下方法向客户添加电子邮件地址

 public class Customer {


    @OneToMany(fetch=FetchType.LAZY, mappedBy="customer", cascade={CascadeType.ALL})
    private List<EmailAddress> emailAddresses;


    public void addEmailAddress(EmailAddress email){
          //As said Hibernate will ignore it when persist this relationship.
          //Add it mainly for the consistency of this relationship for both side in the Java instance 
         this.emailAddresses.add(email);

         email.setCustomer(this);
    }

 }

并且总是调用addeMailAddress()为客户添加电子邮件。您可以将相同的想法应用于更新客户的电子邮件地址。