题
我已经成立了JAX-WS一个SOAP WebServiceProvider此类,但我有麻烦找出如何从为SOAPMessage(或任何节点)对象获取原始的XML。下面是我得到了代码示例,现在,在那里我试图抓住XML:
@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得原始请求的XML?如果有一种方法,通过设置不同类型的供应商来获得原始XML(如源),我愿意做这件事情。
解决方案 3
原来,人们可以通过使用商<源>得到的原始XML,以这种方式:
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;
}
}
我已经结束了不需要这一解决方案,所以我并没有真正尝试过这种代码我自己 - 这可能需要一些调整,以得到正确的。但我知道这是下井得到Web服务的原始XML正确的道路。
(我不知道如何使这项工作,如果你绝对必须有一个SOAPMessage对象,但话又说回来,如果你将要反正处理原始XML,为什么你会使用更高级别的对象? )
其他提示
您可以尝试用这种方法。
SOAPMessage msg = messageContext.getMessage();
ByteArrayOutputStream out = new ByteArrayOutputStream();
msg.writeTo(out);
String strMsg = new String(out.toByteArray());
如果你有SOAPMessage
或SOAPMessageContext
,可以使用一个Transformer
,通过经由Source
其转换为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());
这将编码考虑进去,所以你的“特殊字符”不会得到错位。
有关刚刚调试目的,使用一个行代码 -
msg.writeTo(System.out);
如果您需要格式化XML字符串为XML,试试这个:
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")));
使用变压器厂: -
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();
}
此工作原理
final StringWriter sw = new StringWriter();
try {
TransformerFactory.newInstance().newTransformer().transform(
new DOMSource(soapResponse.getSOAPPart()),
new StreamResult(sw));
} catch (TransformerException e) {
throw new RuntimeException(e);
}
System.out.println(sw.toString());
return sw.toString();
如果您在客户端代码,那么你只需要添加以下两行得到XML请求/响应。这里_call
被org.apache.axis.client.Call
String request = _call.getMessageContext().getRequestMessage().getSOAPPartAsString();
String response = _call.getMessageContext().getResponseMessage().getSOAPPartAsString();
这是很古老的线程,但最近我有一个类似的问题。我打电话下游肥皂服务,从静止服务,我需要返回从下游服务器来为是XML响应。
所以,我最终加入了SOAPMessageContext,处理程序来获取XML响应。然后我注入的响应XML成servlet上下文作为属性。
public boolean handleMessage(SOAPMessageContext context) {
// Get xml response
try {
ServletContext servletContext =
((ServletRequestAttributes) RequestContextHolder.getRequestAttributes()).getRequest().getServletContext();
SOAPMessage msg = context.getMessage();
ByteArrayOutputStream out = new ByteArrayOutputStream();
msg.writeTo(out);
String strMsg = new String(out.toByteArray());
servletContext.setAttribute("responseXml", strMsg);
return true;
} catch (Exception e) {
return false;
}
}
然后,我已经检索到在服务层中的XML响应字符串。
ServletContext servletContext =
((ServletRequestAttributes) RequestContextHolder.getRequestAttributes()).getRequest().getServletContext();
String msg = (String) servletContext.getAttribute("responseXml");
没得机会来测试它尚未但这种方法必须,因为它是使用servlet上下文是线程安全的。