在Java中从SOAPMessage获取原始XML

Question

我在JAX-WS中设置了一个SOAP WebServiceProvider,但是我无法弄清楚如何从SOAPMessage(或任何Node)对象中获取原始XML.这是我现在获得的代码示例,以及我在尝试获取XML的地方:

@WebServiceProvider(wsdlLocation="SoapService.wsdl")
@ServiceMode(value=Service.Mode.MESSAGE)
public class SoapProvider implements Provider<SOAPMessage>
{
    public SOAPMessage invoke(SOAPMessage msg)
    {
        // How do I get the raw XML here?
    }
}

有没有一种简单的方法来获取原始请求的XML？如果有办法通过设置不同类型的Provider(例如Source)来获取原始XML,我也愿意这样做.

Answer 1

你可以试试这种方式.

SOAPMessage msg = messageContext.getMessage();
ByteArrayOutputStream out = new ByteArrayOutputStream();
msg.writeTo(out);
String strMsg = new String(out.toByteArray());

Answer 2

如果您有SOAPMessage或SOAPMessageContext,您可以使用a Transformer,将其转换为Sourcevia DOMSource:

            final SOAPMessage message = messageContext.getMessage();
            final StringWriter sw = new StringWriter();

            try {
                TransformerFactory.newInstance().newTransformer().transform(
                    new DOMSource(message.getSOAPPart()),
                    new StreamResult(sw));
            } catch (TransformerException e) {
                throw new RuntimeException(e);
            }

            // Now you have the XML as a String:
            System.out.println(sw.toString());

这将考虑编码,因此您的"特殊字符"不会被破坏.

Answer 3

事实证明,可以通过使用Provider <Source>以这种方式获取原始XML:

import java.io.ByteArrayOutputStream;
import javax.xml.transform.Source;
import javax.xml.transform.Transformer;
import javax.xml.transform.TransformerFactory;
import javax.xml.transform.stream.StreamResult;
import javax.xml.ws.Provider;
import javax.xml.ws.Service;
import javax.xml.ws.ServiceMode;
import javax.xml.ws.WebServiceProvider;

@ServiceMode(value=Service.Mode.PAYLOAD)
@WebServiceProvider()
public class SoapProvider implements Provider<Source>
{
    public Source invoke(Source msg)
    {
        StreamResult sr = new StreamResult();

        ByteArrayOutputStream out = new ByteArrayOutputStream();
        sr.setOutputStream(out);

        try {
            Transformer trans = TransformerFactory.newInstance().newTransformer();
            trans.transform(msg, sr);

            // Use out to your heart's desire.
        }
        catch (TransformerException e) {
            e.printStackTrace();
        }    

        return msg;
    }
}

我最终不需要这个解决方案,所以我自己没有尝试过这个代码 - 它可能需要一些调整才能正确.但我知道这是从Web服务获取原始XML的正确途径.

(如果你绝对必须拥有一个SOAPMessage对象,我不确定如何使这个工作,但是如果你将要处理原始XML,那么为什么你会使用更高级别的对象呢？)

Answer 4

仅用于调试目的,使用一行代码 -

msg.writeTo(System.out);

Answer 5

如果需要将xml字符串格式化为xml,请尝试以下操作:

String xmlStr = "your-xml-string";
Source xmlInput = new StreamSource(new StringReader(xmlStr));
Transformer transformer = TransformerFactory.newInstance().newTransformer();
transformer.setOutputProperty(OutputKeys.INDENT, "yes");
transformer.setOutputProperty("{http://xml.apache.org/xslt}indent-amount", "2");
transformer.transform(xmlInput,
        new StreamResult(new FileOutputStream("response.xml")));

Answer 6

使用Transformer Factory：

public static String printSoapMessage(final SOAPMessage soapMessage) throws TransformerFactoryConfigurationError,
            TransformerConfigurationException, SOAPException, TransformerException
    {
        final TransformerFactory transformerFactory = TransformerFactory.newInstance();
        final Transformer transformer = transformerFactory.newTransformer();

        // Format it
        transformer.setOutputProperty(OutputKeys.INDENT, "yes");
        transformer.setOutputProperty("{http://xml.apache.org/xslt}indent-amount", "2");

        final Source soapContent = soapMessage.getSOAPPart().getContent();

        final ByteArrayOutputStream streamOut = new ByteArrayOutputStream();
        final StreamResult result = new StreamResult(streamOut);
        transformer.transform(soapContent, result);

        return streamOut.toString();
    }