Question

我有一个XML文件：

<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<object>
   <str>the type</str>
   <bool type="boolean">true</bool>        
</object>

我想把它解组为下面一个类的对象

@XmlRootElement(name="object")
public class Spec  {
   public String str;
   public Object bool;

}

我该怎么做？除非我指定名称空间（见下文），否则它不起作用。

<?xml version="1.0" encoding="UTF-8" standalone="yes"?>
<object>
   <str>the type</str>
   <bool xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"   
       xmlns:xs="http://www.w3.org/2001/XMLSchema"  
       xsi:type="xs:boolean">true</bool>        
</object>

Answer 1

更简单的方法可能是使用unmarshalByDeclaredType，因为您已经知道要解组的类型。

使用

Unmarshaller.unmarshal(rootNode, MyType.class);

您不需要在XML中具有名称空间声明，因为您传入了已设置名称空间的JAXBElement。

这也是完全合法的，因为您不需要在XML实例中引用命名空间，请参阅http://www.w3.org/TR/xmlschema-0/#PO - 许多客户端以这种方式生成XML。

终于开始工作了。请注意，您必须删除架构中的任何自定义命名空间;这是工作示例代码：

架构：

<xsd:schema
   xmlns:xsd="http://www.w3.org/2001/XMLSchema">
<xsd:element name="customer">
   <xsd:complexType>
      <xsd:sequence minOccurs="1" maxOccurs="1">
         <xsd:element name="name" type="xsd:string" minOccurs="1" maxOccurs="1" />
         <xsd:element name="phone" type="xsd:string" minOccurs="1" maxOccurs="1" />
      </xsd:sequence>
   </xsd:complexType>
</xsd:element>

XML：

<?xml version="1.0" encoding="UTF-8"?>
<customer>
   <name>Jane Doe</name>
   <phone>08154712</phone>
</customer>

JAXB代码：

JAXBContext jc = JAXBContext.newInstance(Customer.class);
Unmarshaller u = jc.createUnmarshaller();
u.setSchema(schemaInputStream); // load your schema from File or any streamsource
Customer = u.unmarshal(new StreamSource(inputStream), clazz);  // pass in your XML as inputStream

Answer 2

<强>更新

您可以通过引入中间层在type和xsi:type之间进行转换来实现此目的。下面是使用StAX StreamReaderDelegate为JAXB解组操作执行此操作的示例：

package forum7184526;

import java.io.FileInputStream;

import javax.xml.bind.JAXBContext;
import javax.xml.bind.Unmarshaller;
import javax.xml.stream.XMLInputFactory;
import javax.xml.stream.XMLStreamReader;
import javax.xml.stream.util.StreamReaderDelegate;

import org.eclipse.persistence.oxm.XMLConstants;

public class Demo {

    public static void main(String[] args) throws Exception {
        XMLInputFactory xif = XMLInputFactory.newFactory();
        XMLStreamReader xsr = xif.createXMLStreamReader(new FileInputStream("input.xml"));
        xsr = new XsiTypeReader(xsr);

        JAXBContext jc = JAXBContext.newInstance(Spec.class);
        Unmarshaller unmarshaller = jc.createUnmarshaller();
        Spec spec = (Spec) unmarshaller.unmarshal(xsr);
    }

    private static class XsiTypeReader extends StreamReaderDelegate {

        public XsiTypeReader(XMLStreamReader reader) {
            super(reader);
        }

        @Override
        public String getAttributeNamespace(int arg0) {
            if("type".equals(getAttributeLocalName(arg0))) {
                return XMLConstants.SCHEMA_INSTANCE_URL;
            }
            return super.getAttributeNamespace(arg0);
        }

    }
}

xsi:type是一种用于指定元素实际类型的模式机制（类似于Java中的强制转换）。如果删除命名空间，则表示您正在更改文档的语义。

http://blog.bdoughan.com/2010/11/jaxb-and-inheritance-using-xsitype.html

在EclipseLink JAXB (MOXy)中，我们允许您使用@XmlDescriminatorNode和@XmlDescrimatorValue为域对象指定自己的继承指标。我们目前不提供此类数据类型属性的自定义：

http://blog.bdoughan.com/2010/11/jaxb-and-inheritance-moxy-extension.html

Answer 3

基于Blaise的评论（感谢Blaise！）和我的研究。这是我的问题的解决方案。你同意Blaise，或者你有更好的方法吗？

package forum7184526;

import java.io.FileInputStream;

import javax.xml.bind.JAXBContext;
import javax.xml.bind.Unmarshaller;
import javax.xml.stream.XMLInputFactory;
import javax.xml.stream.XMLStreamReader;
import javax.xml.stream.util.StreamReaderDelegate;

import org.eclipse.persistence.oxm.XMLConstants;

public class Demo {

public static void main(String[] args) throws Exception {
    XMLInputFactory xif = XMLInputFactory.newFactory();
    XMLStreamReader xsr = xif.createXMLStreamReader(new FileInputStream("input.xml"));
    xsr = new XsiTypeReader(xsr);

    JAXBContext jc = JAXBContext.newInstance(Spec.class);
    Unmarshaller unmarshaller = jc.createUnmarshaller();
    Spec spec = (Spec) unmarshaller.unmarshal(xsr);
}

private static class XsiTypeReader extends StreamReaderDelegate {

    public XsiTypeReader(XMLStreamReader reader) {
        super(reader);
    }

    @Override
    public String getAttributeNamespace(int arg0) {
        if("type".equals(getAttributeLocalName(arg0))) {
            return "http://www.w3.org/2001/XMLSchema-instance";
        }
        return super.getAttributeNamespace(arg0);
    }

    @Override
    public String getAttributeValue(int arg0) {
      String n = getAttributeLocalName(arg0);
      if("type".equals(n)) {
         String v = super.getAttributeValue(arg0);
          return  "xs:"+ v;
      }
      return super.getAttributeValue(arg0);
    }
    @Override
    public NamespaceContext getNamespaceContext() {
      return new MyNamespaceContext(super.getNamespaceContext());
    }
}

private static class MyNamespaceContext implements NamespaceContext {
  public NamespaceContext _context;
  public MyNamespaceContext(NamespaceContext c){
     _context = c;
  }

  @Override
  public Iterator<?> getPrefixes(String namespaceURI) {
     return _context.getPrefixes(namespaceURI);
  }

  @Override
  public String getPrefix(String namespaceURI) {
     return _context.getPrefix(namespaceURI);
  }

  @Override
  public String getNamespaceURI(String prefix) {
     if("xs".equals(prefix)) {
        return  "http://www.w3.org/2001/XMLSchema";
     }
     return _context.getNamespaceURI(prefix);
  }
}
}

Answer 4

谢谢大家，在这里分享了适用于我的代码的我的解决方案

我尝试使其通用，每个命名空间都包含“：”，如果有任何标记具有“：”，我将编写代码，它将从xml中删除。

这用于在使用jaxb进行编组时跳过命名空间。

import numpy as npy    
import pandas as pdas     
from matplotlib import pyplot as pt    
import matplotlib    
from sklearn.linear_model import LinearRegression    
import re
from pandasql import sqldf

dataFrame=pdas.read_excel('Project_Airplane_Crashes_and_Fatalities_Since1908.xlsx', sheet="Sheet")
matplotlib.rcParams['figure.figsize'] = (12.0, 8.0)

dataFrame['Year'] = dataFrame['Date'].apply(lambda x: int(str(x)[-4:]))
perYear = dataFrame[['Year','Fatalities']].groupby('Year').agg(['sum','count'])

perYearFig,(xaxis,yaxis)=pt.subplots(2,1,figsize=(15,10))
perYear['Fatalities','sum'].plot(kind='bar',title='Fatalities by Year',grid=True,ax=xaxis,rot=90)
perYear['Fatalities','count'].plot(kind='bar',title='Accidents by Year',grid=True,ax=yaxis,rot=90)
pt.tight_layout()
pt.show()

用于解组，

public class NamespaceFilter {

    private NamespaceFilter() {

    }

    private static final String COLON = ":";

    public static XMLReader nameSpaceFilter() throws SAXException {
        XMLReader xr = new XMLFilterImpl(XMLReaderFactory.createXMLReader()) {
            private boolean skipNamespace;

            @Override
            public void startElement(String uri, String localName, String qName, Attributes atts) throws SAXException {
                if (qName.indexOf(COLON) > -1) {
                    skipNamespace = true;
                } else {
                    skipNamespace = false;
                    super.startElement("", localName, qName, atts);
                }
            }

            @Override
            public void endElement(String uri, String localName, String qName) throws SAXException {
                if (qName.indexOf(COLON) > -1) {
                    skipNamespace = true;
                } else {
                    skipNamespace = false;
                    super.endElement("", localName, qName);
                }
            }

            @Override
            public void characters(char[] ch, int start, int length) throws SAXException {
                if (!skipNamespace) {
                    super.characters(ch, start, length);
                }
            }
        };
        return xr;
    }

}

JAXB：如何在没有命名空间的情况下解组XML

4 个答案: