Hibernate throws org.hibernate.AnnotationException: No identifier specified for entity: com..domain.idea.MAE_MFEView

Question

Why am I getting this exception?

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;
    }
}

Update: I've changed my code to look like this:

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;
    }
}

but now I'm getting this exception:

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

Answer 1

You are missing a field annotated with @Id. Each @Entity needs an @Id - this is the primary key in the database.

If you don't want your entity to be persisted in a separate table, but rather be a part of other entities, you can use @Embeddable instead of @Entity.

If you want simply a data transfer object to hold some data from the hibernate entity, use no annotations on it whatsoever - leave it a simple pojo.

Update: In regards to SQL views, Hibernate docs write:

There is no difference between a view and a base table for a Hibernate mapping. This is transparent at the database level

Answer 2

For me, javax.persistence.Id should be used instead of org.springframework.data.annotation.Id. For anyone who encountered this issue, you can check if you imported the right Id class.

This usually happens if you use IntelliJ's (and perhaps other IDEs too) auto-suggest feature without carefully looking at what the import will be.

Answer 3

This error can be thrown when you import a different library for @Id than Javax.persistance.Id ; You might need to pay attention this case too

In my case I had

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;

when I change the code like this, it got worked

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;

Answer 4

The code below can solve the NullPointerException.

@Id
@GeneratedValue
@Column(name = "STOCK_ID", unique = true, nullable = false)
public Integer getStockId() {
    return this.stockId;
}
public void setStockId(Integer stockId) {
    this.stockId = stockId;
}

If you add @Id, then you can declare some more like as above declared method.

Answer 5

TL;DR

You are missing the @Id entity property, and that's why Hibernate is throwing that exception.

Entity identifiers

Any JPA entity must have an identifier property, that is marked with the Id annotation.

There are two types of identifiers:

• assigned
• auto-generated

Assigned identifiers

An assigned identifier looks as follows:

@Id
private Long id;

Notice that we are using a wrapper (e.g., Long, Integer) instead of a primitive type (e.g., long, int). Using a wrapper type is a better choice when using Hibernate because, by checking if the id is null or not, Hibernate can better determine if an entity is transient (it does not have an associated table row) or detached (it has an associated table row, but it's not managed by the current Persistence Context).

The assigned identifier must be set manually by the application prior to calling persist:

Post post = new Post();
post.setId(1L);

entityManager.persist(post);

Auto-generated identifiers

An auto-generated identifier requires the @GeneratedValue annotation besides the @Id:

@Id
@GeneratedValue
private int id;

There are 3 strategies Hibernate can use to auto-generate the entity identifier:

• IDENTITY
• SEQUENCE
• TABLE

The IDENTITY strategy is to be avoided if the underlying database supports sequences (e.g., Oracle, PostgreSQL, MariaDB since 10.3, SQL Server since 2012). The only major database that does not support sequences is MySQL.

The problem with IDENTITY is that automatic Hibernate batch inserts are disabled for this strategy.

The SEQUENCE strategy is the best choice unless you are using MySQL. For the SEQUENCE strategy, you also want to use the pooled optimizer to reduce the number of database roundtrips when persisting multiple entities in the same Persistence Context.

The TABLE generator is a terrible choice because it does not scale. For portability, you are better off using SEQUENCE by default and switch to IDENTITY for MySQL only.

Answer 6

I think this issue following model class wrong import.

    import org.springframework.data.annotation.Id;

Normally, it should be:

    import javax.persistence.Id;

Answer 7

Using @EmbeddableId for the PK entity has solved my issue.

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

   @EmbeddedId
   SampleEntityPK id;

 }

Answer 8

1. This error occurs due to importing the wrong package:
   import javax.persistence.Id;
2. And you should always give the primary key to the table, otherwise it will give an error.

Answer 9

I know sounds crazy but I received such error because I forget to remove

private static final long serialVersionUID = 1L;

automatically generated by Eclipse JPA tool when a table to entities transformation I've done.

Removing the line above that solved the issue

Answer 10

This error was caused by importing the wrong Id class. After changing org.springframework.data.annotation.Id to javax.persistence.Id the application run

Answer 11

In my case: @Id annotation was: "org.springframework.data.annotation.Id;" After replacing it with that of package: "javax.persistence.Id" the app worked fine.

Answer 12

I had faced same issue while creating session with hibernate using Intelij IDEA. Error was :

org.hibernate.AnnotationException: No identifier specified for entity

This error can caused if @ID annotation is not used in entity class but in my case

Solution Worked for me :

I removed the below import statement from my DAO class

import jakarta.persistence.*;

and added

import javax.persistence.*;

as many users suggested to do .

Answer 13

As somebody has already mentioned the Entity class is missing @Id annotation which is must in case you are persisting your data. Again the @Id must be of same same library as of the @Entity.

Meaning:

javax.persistence.Entity will not work with org.springframework.data.annotation.Id. Both the Entity and Id must either part of Spring Data or Java Persistence API. It is advisable to use Java Persistence API as it is generic.

Answer 14

There can be two areas where we need to look after. i. correct imports ii. missing annotations

annotation required is

import javax.persistence.Id;

and annotation needed is

@Id

Sample Java Class

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;
}

Answer 15

If your table has a combination of fields that make each row unique, you can use a composite key and don't need an id field.

For example

@Entity
@Table(name = "food_nutrient")
@Setter
@Getter
@NoArgsConstructor
@AllArgsConstructor
@ToString
@IdClass(CompositeKey.class)
@EqualsAndHashCode
public class FoodNutrient implements Serializable {

Where CompositeKey is another class like

public class CompositeKey implements Serializable {
    private Integer foodId;
    private Integer nutrientId;

    @Override
    public boolean equals(Object o) {
        if (this == o) return true;
        if (o == null || getClass() != o.getClass()) return false;
        CompositeKey key = (CompositeKey) o;
        if (foodId != key.foodId) return false;
        return nutrientId == key.nutrientId;
    }

    @Override
    public int hashCode() {
        return Objects.hash(foodId, nutrientId);
    }
}

Then your table will have two primary keys with no id field

+--------------+--------------+------+-----+---------+-------+
| Field        | Type         | Null | Key | Default | Extra |
+--------------+--------------+------+-----+---------+-------+
| amount       | float        | YES  |     | NULL    |       |
| fdc_id       | int          | YES  |     | NULL    |       |
| food_id      | int          | NO   | PRI | NULL    |       |
| label_source | varchar(16)  | YES  |     | NULL    |       |
| name         | varchar(255) | YES  |     | NULL    |       |
| nutrient_id  | int          | NO   | PRI | NULL    |       |
| pct_dv       | int          | YES  |     | NULL    |       |
| unit_name    | varchar(16)  | YES  |     | NULL    |       |
+--------------+--------------+------+-----+---------+-------+