Java JAXB带来的XML SOAP信封为空
我试图用JAXB解组XML以将其转换为对象,但是SOAPPArt、SOAPEnvelope和SOAPBody将为空,我不知道为什么 我也尝试过在没有SOAPMessage的情况下解包,但没有成功 以下是我试图解组的XML:Java JAXB带来的XML SOAP信封为空,java,xml,soap,jaxb,unmarshalling,Java,Xml,Soap,Jaxb,Unmarshalling,我试图用JAXB解组XML以将其转换为对象,但是SOAPPArt、SOAPEnvelope和SOAPBody将为空,我不知道为什么 我也尝试过在没有SOAPMessage的情况下解包,但没有成功 以下是我试图解组的XML: <soap:Envelope xmlns:soap="http://schemas.xmlsoap.org/soap/envelope/" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:x
<soap:Envelope xmlns:soap="http://schemas.xmlsoap.org/soap/envelope/"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
<soap:Body>
<ObjectXmlResponse
xmlns="http://tempuri.org/testing">
<ResultadoXml xmlns="www.site.com.br">
<CodigoOperacao>dsadsdas</CodigoOperacao>
<OperacoesObjetoAnalise />
<Respostas>
<Resposta Final="true">
<Sistema>dsadfd</Sistema>
<Criterio>fdsfdsf</Criterio>
<Resposta>.</Resposta>
</Resposta>
</Respostas>
</ResultadoXml>
</ObjectXmlResponse>
</soap:Body>
</soap:Envelope>
public static void main(String[] args) {
JAXBContext jaxbContext;
try {
String relatorio = <the xml>;
InputStream is = new ByteArrayInputStream(relatorio.getBytes());
SOAPMessage message = MessageFactory.newInstance().createMessage(null, is);
SOAPPart sp = message.getSOAPPart();
SOAPEnvelope env = sp.getEnvelope();
SOAPBody bdy = env.getBody();
jaxbContext = JAXBContext.newInstance(ObjectXmlResponse.class);
Unmarshaller jaxbUnmarshaller = jaxbContext.createUnmarshaller();
ObjectXmlResponse response = (ObjectXmlResponse) jaxbUnmarshaller.unmarshal(new StringReader(relatorio));
System.out.println(response);
} catch(Exception ex) {
ex.printStackTrace();
}
System.exit(0);
}
这里是解组:
<soap:Envelope xmlns:soap="http://schemas.xmlsoap.org/soap/envelope/"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
<soap:Body>
<ObjectXmlResponse
xmlns="http://tempuri.org/testing">
<ResultadoXml xmlns="www.site.com.br">
<CodigoOperacao>dsadsdas</CodigoOperacao>
<OperacoesObjetoAnalise />
<Respostas>
<Resposta Final="true">
<Sistema>dsadfd</Sistema>
<Criterio>fdsfdsf</Criterio>
<Resposta>.</Resposta>
</Resposta>
</Respostas>
</ResultadoXml>
</ObjectXmlResponse>
</soap:Body>
</soap:Envelope>
public static void main(String[] args) {
JAXBContext jaxbContext;
try {
String relatorio = <the xml>;
InputStream is = new ByteArrayInputStream(relatorio.getBytes());
SOAPMessage message = MessageFactory.newInstance().createMessage(null, is);
SOAPPart sp = message.getSOAPPart();
SOAPEnvelope env = sp.getEnvelope();
SOAPBody bdy = env.getBody();
jaxbContext = JAXBContext.newInstance(ObjectXmlResponse.class);
Unmarshaller jaxbUnmarshaller = jaxbContext.createUnmarshaller();
ObjectXmlResponse response = (ObjectXmlResponse) jaxbUnmarshaller.unmarshal(new StringReader(relatorio));
System.out.println(response);
} catch(Exception ex) {
ex.printStackTrace();
}
System.exit(0);
}
publicstaticvoidmain(字符串[]args){
JAXBContext JAXBContext;
试一试{
字符串relatorio=;
InputStream is=newbytearrayinputstream(relatorio.getBytes());
SOAPMessage message=MessageFactory.newInstance().createMessage(null,is);
SOAPPart sp=message.getSOAPPart();
SOAPEnvelope env=sp.getEnvelope();
SOAPBody bdy=env.getBody();
jaxbContext=jaxbContext.newInstance(ObjectXmlResponse.class);
解组器jaxbUnmarshaller=jaxbContext.createUnmarshaller();
ObjectXmlResponse-response=(ObjectXmlResponse)jaxbUnmarshaller.unmarshal(新StringReader(relatorio));
System.out.println(响应);
}捕获(例外情况除外){
例如printStackTrace();
}
系统出口(0);
}
我需要填充ObjectXmlResponse对象及其属性,如ResultadoXml。在所有元素上指定名称空间(或在包上使用@XmlSchema),并使用
ObjectXmlResponse response = (ObjectXmlResponse) jaxbUnmarshaller.unmarshal(bdy.extractContentAsDocument());