Question

我用：

mysql-connector-java 6.0.6

hibernate 5.2.10.Final

spring 4.3.8.RELEASE

课程代码：

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);
    }

}

家长班：

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;
    }
}

映射：

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

已经尝试过：

<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>

已经阅读：（所以我希望我不要复制）

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?

还有一些......

错误：

引起：org.hibernate.AnnotationException：没有为实体指定标识符：com.mayan.nst.server.model.PhoneNumber

如果可能我更喜欢解决方案，那么不会生成id

非常感谢你阅读，以及任何帮助

Answer 1

感谢Neil Stocktin问题是我试图从这个孩子那里获得超类属性。（不工作）。解决方案将添加

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

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

并从孩子那里删除

<强>更新

更好的解决方案，将超级儿童桌分开。使用：

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

Hibernate，没有为实体指定标识符，用于映射字符串标识

1 个答案: