Hibernate, No identifier specified for entity, For mapping String id

Question

I use :

mysql-connector-java 6.0.6

hibernate 5.2.10.Final

spring 4.3.8.RELEASE

Class code :

public class PhoneNumber extends AbstractValue{

    public static final int CELL_PHONE = 1, HOME_PHONE = 2, WORK_PHONE = 3;

    public PhoneNumber(String value, Integer type) {
        super(value, type);
    }

    public PhoneNumber() {
        this(null,null);
    }

}

Parent class :

public abstract class AbstractValue {

    private String value;
    private Integer type;

    public AbstractValue(String value, Integer type) {
        this.value = value;
        this.type = type;
    }

    public String getValue() {
        return value;
    }

    public Integer getType() {
        return type;
    }

    public void setValue(String value) {
        this.value = value;
    }

    public void setType(Integer type) {
        this.type = type;
    }
}

Mapping :

<entity class="PhoneNumber" name="PhoneNumber">
        <table name="PhoneNumber"/>
        <attributes>
            <id name="value">
            </id>
            <basic name="type">
            <column nullable="false"/>
        </basic>
    </attributes>
</entity>

Already tried :

<entity class="PhoneNumber" name="PhoneNumber">
    <table name="PhoneNumber"/>
    <attributes>

        <id name="value" access="FIELD">
            <generated-value strategy="SEQUENCE" generator="IdSeq" />
            <sequence-generator name="IdSeq" sequence-name="IdSeq" allocation-size="1" />
        </id>
        <basic name="type">
            <column nullable="false"/>
        </basic>
    </attributes>
</entity>

<entity class="PhoneNumber" name="PhoneNumber">
    <table name="PhoneNumber"/>
    <attributes>

        <id name="value">
            <generated-value strategy="IDENTITY" generator="uuid" />
        </id>
        <basic name="type">
            <column nullable="false"/>
        </basic>
    </attributes>
</entity>

<entity class="PhoneNumber" name="PhoneNumber">
    <table name="PhoneNumber"/>
    <attributes>

        <id name="value">
            <generated-value strategy="TABLE" generator="uuid" />
            <table-generator name="uuid" />
        </id>
        <basic name="type">
            <column nullable="false"/>
        </basic>
    </attributes>
</entity>

And already read : (so i hope i don't do duplicate)

org.hibernate.AnnotationException: No identifier specified for entity: com.ubosque.modelo.Ciudadano

org.hibernate.AnnotationException: No identifier specified for entity: login.Users

java.lang.RuntimeException: org.hibernate.AnnotationException: No identifier specified for entity

Org.Hibernate.AnnotationException: No Identifier Specified For Entity I don't have a id in my table

org.hibernate.AnnotationException: No identifier specified for entity using JPA XML entity-mapping

No Identifier specified exception even when it was

string id generator

How to use @Id with String Type in JPA / Hibernate?

and some more...

error :

Caused by: org.hibernate.AnnotationException: No identifier specified for entity: com.mayan.nst.server.model.PhoneNumber

IF its possible i was prefer a solution were the id will NOT be generated

Thanks you very mutch for read, and for any help

Is AbstractValue marked as an entity or mapped-superclass somewhere? Because it has the field you want to use identity and it isn't currently persistable with what you post there ... — Neil Stockton
– Neil Stockton, Commented Jun 25, 2017 at 12:16
Hey neil thank you for post here, i don't sure i understand you completely. the PhoneNumber is not abstract and the id is regular String. — mayan anger
– mayan anger, Commented Jun 25, 2017 at 12:34
The FIELD that you want to be the "id" is in AbstractValue CLASS. AbstractValue CLASS is NOT marked as "entity" or "mapped-superclass" so you cannot use fields in that class !! Solution is to mark AbstractValue as "mapped-superclass" ... like any JPA doc would tell you — Neil Stockton
– Neil Stockton, Commented Jun 25, 2017 at 12:37
neil stocktin you right ! thank you very so much its work !!!! — mayan anger
– mayan anger, Commented Jun 25, 2017 at 12:49

mayan anger · Accepted Answer · 2017-06-25 13:14:22Z

0

Thanks to Neil Stocktin the problem was that i tried to get superclass property from this child. (not working). solution will be add

<entity class="AbstractValue">
    <attributes>
        <id name="value">

        </id>
    </attributes>
</entity>

and remove the from the child

Update

better solution, to separate the super class childrens tables. use :

<mapped-superclass class="AbstractValue">
    <attributes>
        <id name="value"/>
        <basic name="type">
            <column nullable="false"/>
        </basic>
    </attributes>
</mapped-superclass>

edited Jun 25, 2017 at 13:14

answered Jun 25, 2017 at 12:54

mayan anger

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