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有没有办法使用 JDOM 找到后代元素，就像使用 jQuery 一样？

我正在尝试从此 XML 文档中获取序列元素：

<getEmployeesXMLResponse xmlns="http://tempuri.org/">
    <getEmployeesXMLResult>
        <xs:schema id="NewDataSet" xmlns:xs="http://www.w3.org/2001/XMLSchema"
            xmlns="" xmlns:msdata="urn:schemas-microsoft-com:xml-msdata">
            <xs:element name="NewDataSet" msdata:IsDataSet="true"
                msdata:MainDataTable="ListOfEmployees"
                msdata:UseCurrentLocale="true">
                <xs:complexType>
                    <xs:choice minOccurs="0" maxOccurs="unbounded">
                        <xs:element name="ListOfEmployees">
                            <xs:complexType>
                                <xs:sequence>
                                    <xs:element name="name" type="xs:string" minOccurs="0" />
                                    <xs:element name="department" type="xs:string"
                                        minOccurs="0" />
                                </xs:sequence>
                            </xs:complexType>
                        </xs:element>
                    </xs:choice>
                </xs:complexType>
            </xs:element>
        </xs:schema>
        <diffgr:diffgram xmlns:diffgr="urn:schemas-microsoft-com:xml-diffgram-v1"
            xmlns:msdata="urn:schemas-microsoft-com:xml-msdata">
            <DocumentElement xmlns="">
                <ListOfEmployees diffgr:id="ListOfEmployees1"
                    msdata:rowOrder="0">
                    <name>Paul</name>
                    <department>Sales</department>
                </ListOfEmployees>
                <ListOfEmployees diffgr:id="ListOfEmployees2"
                    msdata:rowOrder="1">
                    <name>Bob</name>
                    <department>Marketing</department>
                </ListOfEmployees>
            </DocumentElement>
        </diffgr:diffgram>
    </getEmployeesXMLResult>
</getEmployeesXMLResponse>

目前我正在这样做：

        Element currentEl = document.getRootElement();
        do {
            currentEl = currentEl.getChildren().get(0);
        } while (currentEl.getName() != "sequence");

但我认为必须有更好的方法。

感谢您的意见

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1 回答 1

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使用 XPaths 并在此处阅读：JavaDoc for XPathFactory以及更多关于它的信息：GitHub Wiki

底线是您希望使用表达式 "//xs:sequence" 和命名空间 Namespace.getNamespace("xs", " http://www.w3.org/2001/XMLSchema ")编译 XPath

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于 2013-04-02T17:31:05.630 回答