在 Java 中从 SOAPMessage 获取原始 XML [英] Getting Raw XML From SOAPMessage in Java

问题描述

我已经在 J​​AX-WS 中设置了一个 SOAP WebServiceProvider，但是我无法弄清楚如何从 SOAPMessage(或任何节点)对象获取原始 XML.这是我现在获得的代码示例，以及我试图获取 XML 的位置:

I've set up a SOAP WebServiceProvider in JAX-WS, but I'm having trouble figuring out how to get the raw XML from a SOAPMessage (or any Node) object. Here's a sample of the code I've got right now, and where I'm trying to grab the 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?
    }
}

有没有简单的方法可以获取原始请求的XML?如果有办法通过设置不同类型的提供程序(例如 Source)来获取原始 XML，我也愿意这样做.

Is there a simple way to get the XML of the original request? If there's a way to get the raw XML by setting up a different type of Provider (such as Source), I'd be willing to do that, too.

推荐答案

原来可以通过使用Provider获取原始XML，方式如下:

It turns out that one can get the raw XML by using Provider<Source>, in this way:

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，所以我自己还没有真正尝试过这个代码 - 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 的正确路径.

I've ended up not needing this solution, so I haven't actually tried this code myself - it might need some tweaking to get right. But I know this is the right path to go down to get the raw XML from a web service.

(如果您绝对必须有一个 SOAPMessage 对象，我不知道如何使这项工作正常进行，但话说回来，如果您无论如何都要处理原始 XML，为什么要使用更高级别的对象?)

(I'm not sure how to make this work if you absolutely must have a SOAPMessage object, but then again, if you're going to be handling the raw XML anyways, why would you use a higher-level object?)

这篇关于在 Java 中从 SOAPMessage 获取原始 XML的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！