Hibernate抛出org.hibernate.AnnotationException:没有为entity指定标识符:com..domain.idea.MAE_MFEView

为什么我得到这个异常?

package com.domain.idea;

import javax.persistence.CascadeType;
import javax.persistence.Entity;
import javax.persistence.FetchType;
import javax.persistence.JoinColumn;
import javax.persistence.OneToOne;
import javax.persistence.Table;

import org.hibernate.annotations.AccessType;

/**
 * object model for the view [InvestmentReturn].[vMAE_MFE]
 */
@Entity
@Table(name="vMAE_MFE", schema="InvestmentReturn")
@AccessType("field")
public class MAE_MFEView
{
    /**
     * trade property is a SuggestdTradeRecommendation object
     */
    @OneToOne(fetch = FetchType.LAZY , cascade = { CascadeType.PERSIST })
    @JoinColumn(name = "suggestedTradeRecommendationID")
    private SuggestedTradeRecommendation trade;

    /**
     * Most Adeverse Excursion value
     */
    private int MAE;

    public int getMAE()
    {
        return MAE;
    }

    /**
     * Most Favorable Excursion value
     */
    private int MFE;

    public int getMFE()
    {
        return MFE;
    }

    /**
     * @return trade property
     * see #trade
     */
    public SuggestedTradeRecommendation getTrade()
    {
        return trade;
    }
}

更新:我修改了我的代码，看起来像这样:

package com.domain.idea;

import javax.persistence.CascadeType;
import javax.persistence.FetchType;
import javax.persistence.Entity;
import javax.persistence.Id;
import javax.persistence.JoinColumn;
import javax.persistence.OneToOne;
import javax.persistence.Table;

import org.hibernate.annotations.AccessType;

/**
 * object model for the view [InvestmentReturn].[vMAE_MFE]
 */
@Entity
@Table(name="vMAE_MFE", schema="InvestmentReturn")
@AccessType("field")
public class MAE_MFEView
{
    /**
     * trade property is a SuggestdTradeRecommendation object
     */
    @Id
    @OneToOne(fetch = FetchType.LAZY , cascade = { CascadeType.PERSIST })
    @JoinColumn(name = "suggestedTradeRecommendationID")
    private SuggestedTradeRecommendation trade;

    /**
     * Most Adeverse Excursion value
     */
    private int MAE;

    public int getMAE()
    {
        return MAE;
    }

    /**
     * Most Favorable Excursion value
     */
    private int MFE;

    public int getMFE()
    {
        return MFE;
    }

    /**
     * @return trade property
     * see #trade
     */
    public SuggestedTradeRecommendation getTrade()
    {
        return trade;
    }
}

但现在我得到这个异常:

Caused by: org.hibernate.MappingException: Could not determine type for: com.domain.idea.SuggestedTradeRecommendation, at table: vMAE_MFE, for columns: [org.hibernate.mapping.Column(trade)]
    at org.hibernate.mapping.SimpleValue.getType(SimpleValue.java:292)
    at org.hibernate.mapping.SimpleValue.isValid(SimpleValue.java:276)
    at org.hibernate.mapping.RootClass.validate(RootClass.java:216)
    at org.hibernate.cfg.Configuration.validate(Configuration.java:1135)
    at org.hibernate.cfg.Configuration.buildSessionFactory(Configuration.java:1320)
    at org.hibernate.cfg.AnnotationConfiguration.buildSessionFactory(AnnotationConfiguration.java:867)
    at org.hibernate.ejb.Ejb3Configuration.buildEntityManagerFactory(Ejb3Configuration.java:669)
    ... 145 more

当前回答

使用@EmbeddableId为PK实体解决了我的问题。

@Entity
@Table(name="SAMPLE")
 public class SampleEntity implements Serializable{
   private static final long serialVersionUID = 1L;

   @EmbeddedId
   SampleEntityPK id;

 }

2019-11-18 04:56:11

其他回答

当您为@Id导入了不同于javax . persistence . id的库时，可能会抛出此错误;你可能也需要注意这个案子

在我的情况下，我有

import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.Table;

import org.springframework.data.annotation.Id;

@Entity
public class Status {

    @Id
    @GeneratedValue
    private int id;

当我像这样修改代码时，它起作用了

import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.Table;

import javax.persistence.Id;

@Entity
public class Status {

    @Id
    @GeneratedValue
    private int id;

2017-12-05 11:06:41

此错误是由导入错误的Id类引起的。将org.springframework.data.annotation.Id更改为javax.persistence.Id后，应用程序将运行

2020-02-29 15:56:46

我们需要关注两个方面。一、正确的进口 2失踪的注释

所需要的注释是

import javax.persistence.Id;

而且所需要的注释是

@Id

Java类示例

package com.oracle.dto;

import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.GeneratedValue;
import javax.persistence.GenerationType;
import javax.persistence.Id;
import javax.persistence.JoinColumn;
import javax.persistence.OneToOne;
import javax.persistence.SequenceGenerator;

import lombok.AccessLevel;
import lombok.Data;
import lombok.Setter;

@Entity
@Data
public class Question {
    @Id
    @Setter(value = AccessLevel.NONE)
    @GeneratedValue(strategy = GenerationType.SEQUENCE,generator = "queGen")
    @SequenceGenerator(name = "queGen",sequenceName = "questionS",initialValue = 1,allocationSize = 1)
    private Integer qId;
    @Column(name="question_text",length = 200)
    private String question;
    @OneToOne
    @JoinColumn(name = "answerId")
    private Answer answer;
}

2022-08-19 07:58:30

正如有人已经提到的实体类缺少@Id注释，这是必须的情况下，你要持久化你的数据。同样，@Id必须与@Entity属于相同的库。

意义:

entity不能使用org.springframework.data.annotation.Id。Entity和Id都必须是Spring Data或Java Persistence API的一部分。建议使用Java Persistence API，因为它是通用的。

2022-06-20 07:05:29

我在使用Intelij IDEA创建hibernate会话时也遇到过同样的问题。错误是:

注解异常:没有为实体指定标识符

如果在实体类中没有使用@ID注释，但在我的情况下，则会导致此错误

解决方法:

我从我的DAO类中删除了下面的导入语句