Java I';我在@OneToMany hibernate映射中出错了?
我试图在两个类之间建立一对多连接,但我遇到了以下错误Java I';我在@OneToMany hibernate映射中出错了?,java,hibernate,spring-boot,annotations,hibernate-mapping,Java,Hibernate,Spring Boot,Annotations,Hibernate Mapping,我试图在两个类之间建立一对多连接,但我遇到了以下错误org.hibernate.AnnotationException:非法尝试将非集合映射为@OneToMany 这是我的密码 Job.java @OneToMany @JoinColumn(name = "id", referencedColumnName = "id", insertable = false, updatable = false) private Set<JobCostSplit> jobCostSplit;
org.hibernate.AnnotationException:非法尝试将非集合映射为@OneToMany
这是我的密码
Job.java
@OneToMany
@JoinColumn(name = "id", referencedColumnName = "id", insertable = false, updatable = false)
private Set<JobCostSplit> jobCostSplit;
您必须将JobCostSplit中的字段注释为@ManyToOne
@ManyToOne
private Job job
您必须将JobCostSplit中的字段注释为@ManyToOne
@ManyToOne
private Job job
您可以使用
@OneToMany
注释您的工作。意味着您将在Job
上有一个属性,在jobcostplit
上有更多属性
因此,如果您想拥有一个作业
和多个作业成本拆分
,您必须将其设置为:
@ManyToOne
@JoinColumn(name = "job_id", referencedColumnName = "id", insertable = false, updatable = false)
private Job job;
您可以使用@OneToMany
注释您的工作。意味着您将在Job
上有一个属性,在jobcostplit
上有更多属性
因此,如果您想拥有一个作业
和多个作业成本拆分
,您必须将其设置为:
@ManyToOne
@JoinColumn(name = "job_id", referencedColumnName = "id", insertable = false, updatable = false)
private Job job;
首先你需要写作
@OneToMany(fetch = FetchType.LAZY, mappedBy = "job")
private Set<JobCostSplit> jobCostSplit= new HashSet<JobCostSplit>();
在JobCostSplit.java中,首先需要编写
@Column(name = "job_id")
private Long jobId;
@JsonIgnore
@OneToMany
@JoinColumn(name = "job_id", referencedColumnName = "id", insertable = false, updatable = false)
private Job job;
@OneToMany(fetch = FetchType.LAZY, mappedBy = "job")
private Set<JobCostSplit> jobCostSplit= new HashSet<JobCostSplit>();
在job.java的jobcostplit.java中用@manytone替换@OneToMany在job.java中用@manytone替换@OneToMany
@Column(name = "job_id")
private Long jobId;
@JsonIgnore
@OneToMany
@JoinColumn(name = "job_id", referencedColumnName = "id", insertable = false, updatable = false)
private Job job;
@ManyToOne
private Job job