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有人可以帮我构建我需要的休眠查询吗？

我有两个表：Areas和AreasLocal。第一个结构为包含应用程序菜单的树表，第二个用作区域名称的 i18n 存储库。

我想根据所选语言（idLocal 属性）尽可能优化整个区域菜单树（急切地使用尽可能少的查询）。

在下一个查询中，我正确绘制了区域树菜单，除了本地化名称，它们看起来是混合的（一些区域名称是西班牙语，其他区域名称是英语等）：

Query query = this.sessionFactory.getCurrentSession().createQuery(
    "select distinct areas" +
    " from Areas areas" +                       
    " left join fetch areas.children childs " +
    " left join fetch areas.locales locales" +
    " where areas.visible = 1 and areas.parent.idArea = :idArea" +  
    " and areasLocal.id.idLocal = :idLocal " +
    " order by areas.orden ");

query.setParameter("idArea", idArea);
query.setParameter("idLocal", idLocal); 




public class Areas implements Serializable {

    private Integer idArea;
    private Areas parent;
    private Integer idTipoAreas;
    private Boolean visible;
    private Integer orden;
    private Set<Areas> children = new HashSet<Areas>(0);
    private Set<AreasLocal> locales = new HashSet<AreasLocal>(0);

    //setters getters ommited       
}

public class AreasLocal implements Serializable {

    private AreasLocalId id;
    private String name;

    //setters getters ommited       
}



public class AreasLocalId implements Serializable {

    private int idArea;
    private int idLocal;

    //setters getters ommited       

}

<hibernate-mapping>
    <class name="Areas" table="areas">
        <id name="idArea" type="java.lang.Integer">
            <column name="ID_AREA" />
        </id>
        <many-to-one name="parent" class="Areas" fetch="select">
            <column name="ID_PARENT"/>
        </many-to-one>
        <property name="idTipoAreas" type="java.lang.Integer">
            <column name="ID_TIPO_AREAS" not-null="true" />
        </property>
        <property name="visible" type="java.lang.Boolean">
            <column name="ES_VISIBLE" not-null="true" sql-type="bit" />
        </property>
        <property name="orden" type="java.lang.Integer">
            <column name="ORDEN" />
        </property>
        <set name="children" table="areas" inverse="true" lazy="false" fetch="select" order-by="ID_AREA">
            <key>
                <column name="ID_PARENT"/>
            </key>
            <one-to-many class="Areas" />
        </set>                
        <set name="locales" table="areas_local" inverse="true" lazy="false" fetch="select">
            <key>
                <column name="ID_AREA"/>
            </key>
            <one-to-many class="AreasLocal" />
        </set>
    </class>
</hibernate-mapping>


<hibernate-mapping>
    <class name="AreasLocal" table="areas_local">
        <composite-id name="id" class="AreasLocalId">
            <key-property name="idArea" type="java.lang.Integer">
                <column name="ID_AREA" />
            </key-property>
            <key-property name="idLocal" type="java.lang.Integer">
                <column name="ID_LOCAL" />
            </key-property>
        </composite-id>
        <property name="name" type="string">
            <column name="NAME" not-null="true" />
        </property>
    </class>
</hibernate-mapping>

我已经意识到HQL 不允许递归连接。

因此，您必须执行以下操作：但也无法按预期工作，并且仅检索具有子项的较深菜单。

Query query = this.sessionFactory.getCurrentSession().createQuery(
    "select distinct areas0" +
    " from Areas areas0" +      

    " left join fetch areas0.children areas1" +
    " left join fetch areas0.locales areas0Locales" +

    " left join fetch areas1.children areas2" +
    " left join fetch areas1.locales areas1Locales" +               

    " left join fetch areas2.children areas3" +
    " left join fetch areas2.locales areas2Locales" +               

    " left join fetch areas3.children areas4" +
    " left join fetch areas3.locales areas3Locales" +               

    //"..." +

    " where areas0.esVisible = 1" +
    " and areas0.parent.idArea = :idArea" +

    " and areas0Locales.id.idLocal = :idLocal " +       
    " and areas1Locales.id.idLocal = :idLocal " +           
    " and areas2Locales.id.idLocal = :idLocal " +           
    " and areas3Locales.id.idLocal = :idLocal " +           

    //"..." +

    " order by areas0.orden ");

任何帮助将不胜感激

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