我有两个实体:Comment
和SubComment
。 Comment
可以有多个SubComment
。我正在尝试与Hibernate建立一对多/多对一的双向关系。
我不知道出了什么问题。这两个表似乎都是在PSQL中正确创建的。
Comment.java
import javax.persistence.*;
import java.util.Set;
@Entity
public class Comment {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private long id;
@Column
private String text;
@OneToMany(cascade = CascadeType.ALL, mappedBy = "comment")
private Set<SubComment> subComment;
public long getId() {
return id;
}
public void setId(long id) {
this.id = id;
}
public String getText() {
return text;
}
public void setText(String text) {
this.text = text;
}
}
SubComment.java
import javax.persistence.*;
@Entity
public class SubComment {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private long id;
private String text;
@ManyToOne
private Comment comment;
public long getId() {
return id;
}
public void setId(long id) {
this.id = id;
}
public String getText() {
return text;
}
public void setText(String text) {
this.text = text;
}
public Comment getComment() {
return comment;
}
public void setComment(Comment comment) {
this.comment = comment;
}
}
我收到了这个错误:
通过JDBC StatementCaused执行DDL时出错: org.postgresql.util.PSQLException:错误:关系“sub_comment”确实 不存在
Hibernate: create table "user" (id bigserial not null, email varchar(255), name varchar(255), username varchar(255), primary key (id))
Hibernate: create table comment (comment_id bigserial not null, text varchar(255), primary key (comment_id))
Hibernate: create table sub_comment (sub_comment_id bigserial not null, text varchar(255), comment_comment_id int8, primary key (sub_comment_id))
Hibernate: alter table sub_comment add constraint FK87789n34vmns9eeyw6jgc5ghp foreign key (comment_comment_id) references comment
application.properties
spring.jpa.hibernate.ddl-auto=create-drop
spring.datasource.url=jdbc:postgresql://localhost:5432/dbname
spring.datasource.data-username=username
spring.datasource.driver-class-name=org.postgresql.Driver
spring.jpa.show-sql=true
spring.jpa.properties.hibernate.dialect=org.hibernate.dialect.PostgreSQLDialect
spring.jpa.database-platform=org.hibernate.dialect.PostgreSQL9Dialect
spring.jpa.properties.hibernate.temp.use_jdbc_metadata_defaults = false
答案 0 :(得分:2)
你错过了@JoinColumn
。由于基于字段的访问,您将收到另一个错误。改为使用基于属性的访问:
import javax.persistence.*;
@Entity
@Table(name = "subcomment")
public class SubComment implements Serializable {
private static final long serialVersionUID = -3009157732242241606L;
private long id;
private String text;
private Comment comment;
@Id
@Column(name = "sub_id")
@GeneratedValue(strategy = GenerationType.AUTO)
public long getId() {
return id;
}
public void setId(long id) {
this.id = id;
}
@Basic
@Column(name = "sub_text")
public String getText() {
return text;
}
public void setText(String text) {
this.text = text;
}
@ManyToOne
@JoinColumn(name = "sub_fk_c_id", referencedColumnName = "c_id") // here the exact field name of your comment id in your DB
public Comment getComment() {
return comment;
}
public void setComment(Comment comment) {
this.comment = comment;
}
}
此处也进行更改:
import javax.persistence.*;
@Entity
@Table(name = "comment")
public class Comment implements Serializable {
private static final long serialVersionUID = -3009157732242241606L;
private long id;
private String text;
private Set<SubComment> subComment = new HashSet<>();
@OneToMany(mappedBy = "comment", targetEntity = SubComment.class)
public Set<SubComment> getSubComment() {
return subComment;
}
public void setSubComment(Set<SubComment> subComment) {
this.subComment = subComment;
}
@Id
@Column(name = "c_id")
@GeneratedValue(strategy = GenerationType.AUTO)
public long getId() {
return id;
}
public void setId(long id) {
this.id = id;
}
@Basic
@Column(name = "c_text")
public String getText() {
return text;
}
public void setText(String text) {
this.text = text;
}
}
将以下内容粘贴到application.properties
文件中:
spring.jpa.properties.hibernate.id.new_generator_mappings = false
spring.jpa.properties.hibernate.format_sql = true
logging.level.org.hibernate.SQL=DEBUG
logging.level.org.hibernate.type.descriptor.sql.BasicBinder=TRACE
spring.jpa.hibernate.ddl-auto=create
在你的pom.xml文件中粘贴这些:
<dependency>
<groupId>org.postgresql</groupId>
<artifactId>postgresql</artifactId>
<scope>runtime</scope>
</dependency>
有关详细信息,请参阅this stackoverflow post。
答案 1 :(得分:1)
我知道这个答案来晚了,但是也许会对某人有所帮助,他们的问题是他们没有配置架构,这就是为什么抛出该错误的原因。
以下是在postgresql中配置架构的方法:
spring.jpa.properties.hibernate.default_schema = "his schema"
答案 2 :(得分:0)
许多子评论属于一个评论,模型看起来应该是这样(例如,您可以生成例如UUID.randomUUID()。toString())
@Entity
@Table(name = "comment")
public class Comment{
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
@Column(name = "comment_id")
String commentId;
}
@Entity
@Table(name = "sub_comment")
public class SubComment{
...
@ManyToOne(fetch = FetchType.LAZY)
@JoinColumn(name = "comment_id", referencedColumnName = "comment_id")
Comment comment;
...
}
如果您使用例如liquidbase来控制数据源:
- changeSet:
id: 1
author: author.name
changes:
- addForeignKeyConstraint:
baseColumnNames: comment_id
baseTableName: sub_comment
constraintName: fk_comment_subcomment
referencedColumnNames: comment_id
referencedTableName: comment