我正在使用 JAXB 来解析 xml。我有一个如下架构,以及在此架构上定义的两个 xml 文件 a.xml 和 b.xml。a.xml 通过 xi:include xml 标记依赖于 b.xml。请归档以下示例以获得更清晰的数据
I have followng schema definition:
<?xml version="1.0" encoding="UTF-8"?>
<xs:schema xmlns:xs="http://www.w3.org/2001/XMLSchema" elementFormDefault="qualified" attributeFormDefault="unqualified">
<xs:element name="Task">
<xs:complexType>
<xs:sequence>
<xs:element ref="Details" minOccurs="1" maxOccurs="unbounded"/>
</xs:sequence>
</xs:complexType>
</xs:element>
<xs:element name="Details">
<xs:complexType>
<xs:sequence>
<xs:element name="NAme" type="xs:string"/>
</xs:sequence>
</xs:complexType>
</xs:element>
</xs:schema>
这是两个xml文件:
一个.xml:
<?xml version="1.0" encoding="UTF-8"?>
<Task xmlns:xi="http://www.w3.org/2001/XInclude">
<Details>
<xi:include href="b.xml"/>
</Details>
</Task>
b.xml:
<?xml version="1.0" encoding="UTF-8"?>
<Detail>
<Name>Name1</Name>
</Detail>
<Detail>
<Name>Name2</Name>
</Detail>
现在我使用 JAXB SAXFactory 将其解析为:
JAXBContext jaxbcon = JAXBContext.newInstance("schema-definition-jaxb-files");
unmar = jaxbcon.createUnmarshaller();
SAXParserFactory spf = SAXParserFactory.newInstance();
spf.setXIncludeAware(true);
XMLReader xr = spf.newSAXParser().getXMLReader();
SAXSource source = new SAXSource(xr, new InputSource(new FileInputStream(xmlfilename)));
Object obj = unmar.unmarshal(source);
解析成功,但Details JAXB 标记对象为空。无论如何,a.xml 文件中的 xi:include 标记没有被展平。任何想法?