XQUERY JAVA不工作

我从一个复杂的XML文件中获取一些属性：

<rsp>
    <csl  d='10775.916760613756' id='2003' nam='AUTOS TEZIUTLÁN, S.A. DE C.V.' 
          adr='KM. 1 CARR. TEZIUTLÁN-TLAPACOYAN' tel='231312-12-05' 
          lat='19.826765' lon='-97.347906' />
    <csl  d='10789.680721293766' id='2019' nam='AUTOMOVILÍSTICA DE TEHUACAN, S.A. DE C.V.' 
          adr='BLVD. ADOLFO LOPEZ MATEOS NO. 3623' tel='238382-44-33' 
          lat='18.467281' lon='-97.417901' />
    <csl  d='10848.586325071066' id='2013' nam='AUTOMOTRIZ DE LA SIERRA, S.A. DE C.V.' 
          adr='AUSENCIO T. JIMÉNEZ No. 1' tel='776762-05-42' 
          lat='20.174386' lon='-98.06125' />
    <csl  d='10866.815936520663' id='2028' nam='MOTORES ALEMANES RIVERA S.A. DE C.V.' 
          adr='CALZADA IGNACIO ZARAGOZA NO. 180' tel='222286-02-02' 
          lat='19.064258' lon='-98.179042' />
    <csl  d='10867.374198658401' id='2012' nam='ARMENTA AUTOMOTRIZ, S.A. DE C.V.' 
          adr='24 NORTE No. 214' tel='222235-87-68' 
          lat='19.038912' lon='-98.183101' />
</rsp>

---

我一直在使用其他查询，但结果总是第一个元素。。。我可能缺少什么？

XPath将返回文档中所有

@d

属性节点的列表

您的Java代码：

  else if (result instanceof NodeList)    {
        NodeList list = (NodeList) result;

        Node node = list.item(0); // <-- THIS ONE HERE

        Log.d("LIST", Integer.toString(list.getLength()));
        ret = node.getTextContent();
    }

else if（节点列表的结果实例）{
节点列表=（节点列表）结果；
Node Node=list.item（0）；//请注意，您没有使用XQuery，而是使用XPath 1.0。
String loc1 = XMLParser.getParamByXPath(service, "//@d");

  else if (result instanceof NodeList)    {
        NodeList list = (NodeList) result;

        Node node = list.item(0); // <-- THIS ONE HERE

        Log.d("LIST", Integer.toString(list.getLength()));
        ret = node.getTextContent();
    }