Java 子表JPA的持久化问题
我在MySQl中有一个OneToMany关系,我使用JPA映射它。从数据库中检索信息时,它工作得很好,但在持久化子表时,我遇到了一个问题。 --MySQL工作台正向工程Java 子表JPA的持久化问题,java,mysql,spring,hibernate,spring-data-jpa,Java,Mysql,Spring,Hibernate,Spring Data Jpa,我在MySQl中有一个OneToMany关系,我使用JPA映射它。从数据库中检索信息时,它工作得很好,但在持久化子表时,我遇到了一个问题。 --MySQL工作台正向工程 SET @OLD_UNIQUE_CHECKS=@@UNIQUE_CHECKS, UNIQUE_CHECKS=0; SET @OLD_FOREIGN_KEY_CHECKS=@@FOREIGN_KEY_CHECKS, FOREIGN_KEY_CHECKS=0; SET @OLD_SQL_MODE=@@SQL_MODE, SQL_M
SET @OLD_UNIQUE_CHECKS=@@UNIQUE_CHECKS, UNIQUE_CHECKS=0;
SET @OLD_FOREIGN_KEY_CHECKS=@@FOREIGN_KEY_CHECKS, FOREIGN_KEY_CHECKS=0;
SET @OLD_SQL_MODE=@@SQL_MODE, SQL_MODE='ONLY_FULL_GROUP_BY,STRICT_TRANS_TABLES,NO_ZERO_IN_DATE,NO_ZERO_DATE,ERROR_FOR_DIVISION_BY_ZERO,NO_ENGINE_SUBSTITUTION';
-- -----------------------------------------------------
-- Schema mydb
-- -----------------------------------------------------
-- -----------------------------------------------------
-- Schema mydb
-- -----------------------------------------------------
CREATE SCHEMA IF NOT EXISTS `mydb` DEFAULT CHARACTER SET utf8 ;
USE `mydb` ;
-- -----------------------------------------------------
-- Table `mydb`.`course`
-- -----------------------------------------------------
CREATE TABLE IF NOT EXISTS `mydb`.`course` (
`id` INT NOT NULL AUTO_INCREMENT,
`name` VARCHAR(45) NULL,
`descr` VARCHAR(45) NULL,
PRIMARY KEY (`id`))
ENGINE = InnoDB;
-- -----------------------------------------------------
-- Table `mydb`.`topic`
-- -----------------------------------------------------
CREATE TABLE IF NOT EXISTS `mydb`.`topic` (
`id` INT NOT NULL AUTO_INCREMENT,
`name` VARCHAR(45) NULL,
`descr` VARCHAR(45) NULL,
`course_id` INT NOT NULL,
PRIMARY KEY (`id`, `course_id`),
INDEX `fk_topic_course_idx` (`course_id` ASC) VISIBLE,
CONSTRAINT `fk_topic_course`
FOREIGN KEY (`course_id`)
REFERENCES `mydb`.`course` (`id`)
ON DELETE NO ACTION
ON UPDATE NO ACTION)
ENGINE = InnoDB;
SET SQL_MODE=@OLD_SQL_MODE;
SET FOREIGN_KEY_CHECKS=@OLD_FOREIGN_KEY_CHECKS;
SET UNIQUE_CHECKS=@OLD_UNIQUE_CHECKS;
@Entity
@Table(name = "course", schema = "mydb")
public class Course implements Serializable {
/**
*
*/
private static final long serialVersionUID = 1886462862094528507L;
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Column(name = "id")
private int id;
@Column(name = "name")
private String name;
@Column(name = "descr")
private String descr;
@OneToMany(mappedBy = "course", fetch = FetchType.EAGER) //name of object in Topic entity
@JsonManagedReference
private Set<Topic> topic;
}
@Entity
@Table(name="topic", schema = "mydb")
public class Topic implements Serializable {
/**
*
*/
private static final long serialVersionUID = -365302645107748753L;
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Column(name = "id", updatable = false)
private int id;
@Column(name = "name")
private String name;
@Column(name = "descr")
private String descr;
@ManyToOne(cascade=CascadeType.ALL)
@JoinColumn(name="course_id") //name of the foreign key. Must pay attention here
@JsonBackReference
private Course course;
}
2019-11-14 13:11:27.013错误18916---[nio-8081-exec-2]o.a.c.c.c.[/].[dispatcherServlet]:路径为[]的上下文中Servlet[dispatcherServlet]的Servlet.service()引发异常[请求处理失败;嵌套异常为org.springframework.dao.DataIntegrityViolationException:无法执行语句;SQL[n/a];约束[null];嵌套异常是org.hibernate.exception.ConstraintViolationException:无法执行语句],其根本原因是
java.sql.SQLIntegrityConstraintViolationException:列“课程id”不能为null
出现此错误是因为您使用的是
@GeneratedValue
,没有附加规范。默认情况下,Hibernate将尝试访问名为。Hibernate sequence
的序列
由于您使用的是mysql,因此可以注释:
@GeneratedValue(strategy = GenerationType.IDENTITY)
这将导致hibernate使用mysql的AUTO_INCREMENT
而不是sequence
您还需要在创建架构时使用AUTO_INCREMENT
定义列
在持久化之前,您还缺少对课程
实体的引用:
topic.setCourse(entityManager.getReference(Course.class, courseId));
courseId
这是作为json传递的course\u id
。这是怎么回事?一个实体,这也是一个存储库?我强烈建议至少读一读jpa。Andronicus,@repository注释是错误的。谢谢你的建议,我已经删除了它。这些注释不是用于扩展讨论的这段对话已经结束了。
topic.setCourse(entityManager.getReference(Course.class, courseId));