gpt4 book ai didi

java - Spring JPA Crud 存储库保存不返回 UUID 字段

转载 作者:行者123 更新时间:2023-11-29 12:40:34 25 4
gpt4 key购买 nike

我尝试使用纯 JPA 保存实体,但保存时未返回 Postgres UUID 字段。默认情况下,UUID 字段是使用 public.uuid_generate_v1mc() 自动生成的。我使用 postgres 作为数据库,由于某种原因,在 jpa 中保存时创建的默认值不会返回。我尝试添加一个 @Generate 字段并将其设置为自动,但这没有任何作用。

我做错了什么吗?

数据库创建语句

CREATE TABLE usermgmt.test
(
id SERIAL PRIMARY KEY,
active boolean NOT NULL DEFAULT false,
created timestamp without time zone NOT NULL DEFAULT now(),
updated timestamp without time zone NOT NULL DEFAULT now(),
description character varying,
name character varying NOT NULL,
external_id bigint UNIQUE ,
uuid uuid DEFAULT public.uuid_generate_v1mc(),
alias character varying NOT NULL UNIQUE,
code character varying NOT NULL UNIQUE
);

实体

@SuppressWarnings("serial")
@Entity(name = "managementTest")
@Cacheable
@Table(schema = "usermgmt", name = "test")
public class Test extends BaseEntity implements Serializable {

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

@Column(nullable = false)
private String name;

@Column(nullable = false)
private String description;

@Column(nullable = false, insertable = false, updatable = false)
private UUID uuid;

@Column(nullable = false, unique = true)
private String alias;

@Column(nullable = false, unique = true)
private String code;

public Long getId() {
return id;
}

public void setId(Long id) {
this.id = id;
}

/**
* @return the name
*/
public String getName() {
return name;
}

/**
* @param name the name to set
*/
public void setName(String name) {
this.name = name;
}

/**
* @return
*/
public String getDescription() {
return description;
}

/**
* @param description
*/
public void setDescription(String description) {
this.description = description;
}

/**
* @return the uuid
*/
public UUID getUuid() {
return uuid;
}

public void setUuid(UUID uuid) {
this.uuid = uuid;
}

public String getAlias() {
return alias;
}

public void setAlias(String alias) {
this.alias = alias;
}

public String getCode() {
return code;
}

public void setCode(String code) {
this.code = code;
}
}

基础实体

@MappedSuperclass
public abstract class BaseEntity {
private LocalDateTime created;
private LocalDateTime updated;
@NotNull
private Boolean active;

public BaseEntity() {
this.active = Boolean.TRUE;
}

@PrePersist
@PreUpdate
void setDates() {
if (this.created == null) {
this.created = LocalDateTime.now();
}

this.updated = LocalDateTime.now();
}

public Boolean getActive() {
return this.active;
}

public void setActive(Boolean active) {
this.active = active;
}

public LocalDateTime getCreated() {
return this.created;
}

public LocalDateTime getUpdated() {
return this.updated;
}

public void setUpdated(LocalDateTime updated) {
this.updated = updated;
}

public void touch() {
this.updated = LocalDateTime.now();
}
}

原油

@Repository("managementTestCrudRepository")
public interface TestCrudRepository extends JpaRepository<Test, Long> {
Test findByCode(String code);
Test findByUuid(UUID uuid);
Test findByAlias(String alias);
List<Test> findByActive(Boolean active);
}

这是我用来保存实体的方法

@PutMapping
public Test putByJson(@RequestBody String json) {
return testCrudRepository.save(new Gson().fromJson(json, Test.class));
}

postman 中返回的对象

{
"created": {
"month": "OCTOBER",
"year": 2018,
"dayOfYear": 275,
"hour": 11,
"minute": 9,
"nano": 325000000,
"second": 52,
"dayOfMonth": 2,
"dayOfWeek": "TUESDAY",
"monthValue": 10,
"chronology": {
"id": "ISO",
"calendarType": "iso8601"
}
},
"updated": {
"month": "OCTOBER",
"year": 2018,
"dayOfYear": 275,
"hour": 11,
"minute": 9,
"nano": 329000000,
"second": 52,
"dayOfMonth": 2,
"dayOfWeek": "TUESDAY",
"monthValue": 10,
"chronology": {
"id": "ISO",
"calendarType": "iso8601"
}
},
"active": true,
"id": 2,
"externalId": null,
"name": "test1",
"description": "test1",
"uuid": null,
"alias": "test1",
"code": "test1"
}

最佳答案

解决方案

您需要添加 @Generated对数据库将生成的字段进行注释。

@Generated(GenerationTime.ALWAYS)
private UUID uuid;

说明

您需要找到一种方法来告诉您的 JPA 提供商 (1) 在数据库端生成的字段以及 (2) 何时获取它们。

对于此类属性,有一个名为 @Generated 的注释。它提供了两个选项:

  • 在插入时获取生成的属性 (GenerationTime.INSERT);
  • 在插入和更新时获取生成的属性 (GenerationTime.ALWAYS)。

为了检索所需的 @Generate 值,Hibernate(Spring 将其用作默认 JPA 提供程序)将执行附加的 SELECT 语句。

我不知道您将使用 PUT 端点来做什么(创建/更新/覆盖资源1),但我会选择 @Generate(ALWAYS) 因为,使用 @Generate(INSERT),您在更新/合并时将得到 null

奖金

我发现这个视频内容丰富且切题。

video


1 a serious discussion here

关于java - Spring JPA Crud 存储库保存不返回 UUID 字段,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/52612241/

25 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com