如何使用XPath找到xml中属性为ref的节点？？

Question

 <xs:attributeGroup name="arcAttrs">
        <xs:attribute ref="xlink:type" use="required" fixed="arc"/>
        <xs:attribute ref="xlink:arcrole"/>
        <xs:attribute ref="xlink:title"/>
        <xs:attribute ref="xlink:show"/>
        <xs:attribute ref="xlink:actuate"/>
        <xs:attribute ref="xlink:from"/>
        <xs:attribute ref="xlink:to">
            <xs:annotation>
                <xs:documentation>
     from and to have default behavior when values are missing
    </xs:documentation>
            </xs:annotation>
        </xs:attribute>
    </xs:attributeGroup>
    <xs:group name="arcModel">
        <xs:sequence>
            <xs:element ref="xlink:title" minOccurs="0" maxOccurs="unbounded"/>
        </xs:sequence>
    </xs:group>
    <xs:complexType name="arcType">
        <xs:group ref="xlink:arcModel"/>
        <xs:attributeGroup ref="xlink:arcAttrs"/>
    </xs:complexType>

类似于上面这段代码，怎样使用XPath获得 或
？

做这个是想要使用java类的xjc将xsd直接生成为java包，但这里有两个xlink:title 出现明明冲突的问题，因此想通过xjb文件来重新定义title名称，but一直取不到这两个节点，求高手相助

Answer 1

https://q.cnblogs.com/q/84028/