如何使用java解析XOP/MTOM SOAP响应

Question

我只想知道，有没有解析MTOM/XOP SOAP响应的简单方法。问题是，我使用纯HTTP发送soap消息，并使用javax.xml解析响应。但是有些服务用mulipart/related来响应我，并且需要更复杂的逻辑来解析它（性能很重要）。所以我想知道我是否可以以某种方式利用apachecxf、apacheaxiom或任何其他库来解析MTOM/XOP SOAP响应？

Answer 1

这些单元测试向您展示了如何使用CXF从MTOM消息中提取附件。我将内联其中一个测试，以防将来不存在此链接：

private MessageImpl msg;
@Before
public void setUp() throws Exception {
    msg = new MessageImpl();
    Exchange exchange = new ExchangeImpl();
    msg.setExchange(exchange);
}
@Test
public void testDeserializerMtom() throws Exception {
    InputStream is = getClass().getResourceAsStream("mimedata");
    String ct = "multipart/related; type="application/xop+xml"; "
                + "start="<[e-mail]>"; "
                + "start-info="text/xml; charset=utf-8"; "
                + "boundary="----=_Part_4_701508.1145579811786"";
    msg.put(Message.CONTENT_TYPE, ct);
    msg.setContent(InputStream.class, is);
    AttachmentDeserializer deserializer = new AttachmentDeserializer(msg);
    deserializer.initializeAttachments();
    InputStream attBody = msg.getContent(InputStream.class);
    assertTrue(attBody != is);
    assertTrue(attBody instanceof DelegatingInputStream);
    Collection<Attachment> atts = msg.getAttachments();
    assertNotNull(atts);
    Iterator<Attachment> itr = atts.iterator();
    assertTrue(itr.hasNext());
    Attachment a = itr.next();
    assertNotNull(a);
    InputStream attIs = a.getDataHandler().getInputStream();
    // check the cached output stream
    ByteArrayOutputStream out = new ByteArrayOutputStream();
    IOUtils.copy(attBody, out);
    assertTrue(out.toString().startsWith("<env:Envelope"));
    // try streaming a character off the wire
    assertTrue(attIs.read() == '/');
    assertTrue(attIs.read() == '9');
}

在您的情况下，ct将来自响应的内容类型标头。"mimedata"将是响应的内容。

Answer 2

无需使用CXF，标准的javax.mail.internet.MimeMultipart类可以完成这项工作，而且非常容易使用（也可以创建MTOM请求）。

这里有一个非常简单的例子来解码MTOM响应的部分：

MimeMultipart mp = new MimeMultipart(new ByteArrayDataSource(data, contentType));
int count = mp.getCount();
for (int i = 0; i < count; i++) {
    BodyPart bp = mp.getBodyPart(i);
    bp.saveFile(filepath + "_" + i);
}

Answer 3

我遇到了与@Nicolas Albert 相同的问题并得到了解决

public byte[] mimeParser(InputStream isMtm) {
    ByteArrayOutputStream baos = null;
    try {
        MimeMultipart mp = new MimeMultipart(new ByteArrayDataSource(isMtm,
                ct));
        int count = mp.getCount();
        baos = new ByteArrayOutputStream();
        for (int i = 0; i < count; i++) {
            BodyPart bodyPart = mp.getBodyPart(i);
            if (!Part.ATTACHMENT
                    .equalsIgnoreCase(bodyPart.getDisposition())
                    && !StringUtils.isNotBlank(bodyPart.getFileName())) {
                continue; // dealing with attachments only
            }
            bodyPart.writeTo(baos);
        }
        byte[] attachment = baos.toByteArray();
        FileUtils.writeByteArrayToFile(new File("E:/wss/attachment.zip"), attachment);
        return attachment;
    } catch (Exception ex) {
        ex.printStackTrace();
    } finally {
        if (baos != null) {
            try {
                baos.close();
            } catch (Exception ex) {
            }
        }
    }
    return null;
}