Spring Boot & Postgres：關系“sub_comment”不存在

Question

我有兩個實體： Comment和SubComment 。 一個Comment可以有多個SubComment 。 我正在嘗試與 Hibernate 建立一對多/多對一的雙向關系。

我不知道出了什么問題。 這兩個表似乎都是在 PSQL 中正確創建的。

注釋.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;
    }
}

子注釋.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;
    }
}

我收到此錯誤：

通過 JDBC 語句執行 DDL 時出錯引起：org.postgresql.util.PSQLException：錯誤：關系“sub_comment”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

應用程序屬性

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

Answer 1

你錯過了@JoinColumn 。 由於基於字段的訪問，您將收到另一個錯誤。 改用基於屬性的訪問：

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;
    }
}

也在這里進行更改：

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;
    }
}

將以下內容粘貼到您的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 文件中粘貼這些：

    <dependency>
        <groupId>org.postgresql</groupId>
        <artifactId>postgresql</artifactId>
        <scope>runtime</scope>
    </dependency>

如需進一步參考，請參閱此 stackoverflow 帖子。

Answer 2

我知道這個答案來晚了，但也許它會幫助某人，他們的問題是他們沒有配置架構，這就是他們拋出該錯誤的原因。

以下是如何在 postgresql 中配置架構：

spring.jpa.properties.hibernate.default_schema = "his schema"

Answer 3

許多子評論屬於一個評論，模型應該看起來像這樣（comment_id 你可以生成例如 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 來控制您的數據源：

  - changeSet:
      id: 1
      author: author.name
      changes:
      - addForeignKeyConstraint:
          baseColumnNames: comment_id
          baseTableName: sub_comment
          constraintName: fk_comment_subcomment
          referencedColumnNames: comment_id
          referencedTableName: comment

Answer 4

有一個類似的問題，我的問題是我的 SubCategory 類下缺少 referencedColumnName。

有這個

  @ManyToOne(fetch = FetchType.LAZY, cascade = CascadeType.MERGE)
    @JoinColumn(name = "category_id, nullable = false)
    private Category category;

代替

@ManyToOne(fetch = FetchType.LAZY, cascade = CascadeType.MERGE)
    @JoinColumn(name = "category_id", referencedColumnName = "id", nullable = false)
    private Category category;

所以我的子類別表沒有被創建。