Mule 如何按名称获取所有XML节点？

我有这样的输入XML-

<parent>
    <child type="reference">
        <grandChild name="aaa" action="None">
            <Attribute name="xxx">1</Attribute>
            <grandChild name="bbb" action="None">
                <Attribute name="xxx">1</Attribute>
            </grandChild>
            <grandChild name="aaa" action="None">
                <Attribute name="xxx">2</Attribute>
            </grandChild>
        </grandChild>
        <grandChild name="ddd" action="None">
                <Attribute name="xxx">1</Attribute>
                <grandChild name="aaa" action="None">
                    <Attribute name="xxx">3</Attribute>
                </grandChild>
        </grandChild>
    </child>
</parent>

---

这将使用..*选择器返回所需的输出，以检索所有子项并重建输出结构：

%dw 2.0
output application/xml
---

grandChilds:{
    ( payload.parent..*grandChild filter($.@name == 'aaa') map(gc) ->{

    grandChild @(name: gc.@name, action: gc.@action): {
        Attribute @(name: gc.Attribute.@name): gc.Attribute
    }

})

}

产出：

<?xml version='1.0' encoding='UTF-8'?>
<grandChilds>
  <grandChild name="aaa" action="None">
    <Attribute name="xxx">1</Attribute>
  </grandChild>
  <grandChild name="aaa" action="None">
    <Attribute name="xxx">2</Attribute>
  </grandChild>
  <grandChild name="aaa" action="None">
    <Attribute name="xxx">3</Attribute>
  </grandChild>
</grandChilds>

<?xml version='1.0' encoding='UTF-8'?>
<grandChilds>
  <grandChild name="aaa" action="None">
    <Attribute name="xxx">1</Attribute>
  </grandChild>
  <grandChild name="aaa" action="None">
    <Attribute name="xxx">2</Attribute>
  </grandChild>
  <grandChild name="aaa" action="None">
    <Attribute name="xxx">3</Attribute>
  </grandChild>
</grandChilds>