Question

我正在尝试解组与JAXB类不匹配的XML字符串。我希望这会抛出一个异常，但是会返回一个新的JAXB对象。

xmlStr（输入XML）

<urn1:RgBad
    xmlns:urn1="urn:none">
</urn1:RgBad>

更正XML

<urn:Rg
    xmlns:urn="urn:test"

代码

(clazz = Rg.class)
    JAXBContext jaxbContext = JAXBContext.newInstance(clazz);
    Unmarshaller unmarshaller = jaxbContext.createUnmarshaller();
    StreamSource source = new StreamSource(new StringReader(xmlStr));
    //Returns an actual Rg object, even though the source root element and namespace are different.
    (unmarshaller.unmarshal(source, clazz)).getValue();

Answer 1

您可以尝试添加架构以在使用JAXB时验证xml文件

try {  
  SchemaFactory sf = SchemaFactory.newInstance(XMLConstants.W3C_XML_SCHEMA_NS_URI);  
  Schema schema = sf.newSchema(new  File( "yourSchema.xsd" ));  
  JAXBContext jc = JAXBContext.newInstance(clazz.getPackage().getName());  
  Unmarshaller u = jc.createUnmarshaller();  
  u.setSchema(schema);  
  u.setEventHandler(ehdler);  

  obj = u.unmarshal(xml);  
} catch  (JAXBException e) {
} finally  {  
}

JAXB解组错误的XML不会引发异常

1 个答案: