我有一个实现 SOAPHandler interface 的类. handleMessage 定义为:

public boolean handleMessage(SOAPMessageContext context) {

  SOAPMessage msg = context.getMessage();
  SOAPPart part = msg.getSOAPPart();
  SOAPEnvelope envelope = part.getEnvelope();

  // add namespaces
  SOAPElement envelope.addNamespaceDeclaration("xsd", "http://www.w3.org/2001/XMLSchema");
  envelope.addNamespaceDeclaration("xsi", "http://www.w3.org/2001/XMLSchema-      

  // add the header with additional elements
  Name qname = envelope.createName("Security", "sse", "http://example.com/security.xsd");
  element = envelope.addHeader().addChildElement(qname);

  qname = envelope.createName("mustUnderstand");
  element.addAttribute(qname, "1");

  qname = envelope.createName("UsernameToken", "sse", "http://example.com/user.xsd");
  element = envelope.getHeader().addHeaderElement(qname);
  element.addTextNode("user1");

  qname = envelope.createName("Password");
  element = envelope.getHeader().addHeaderElement(qname);
  element.addTextNode("1234");

}

} catch (Exception e) {
  e.printStackTrace();
}
  return true;
}

这会生成以下消息:

<S:Envelope xmlns:S="http://schemas.xmlsoap.org/soap/envelope/" xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
  <S:Header>
    <sse:Security xmlns:sse="http://example.com/security.xsd" mustUnderstand="1"/>
    <sse:UsernameToken xmlns:sse="http://example.com/user.xsd">user1</sse:UsernameToken>
  </S:Header>
  <S:Body>
    ....The rest of the transaction
  </S:Body>
</S:Envelope>

问题是我需要生成具有以下格式的消息:

<soapenv:Envelope xmlns:soapenv="http://schemas.xmlsoap.org/soap/envelope/" xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
   <soapenv:Header>
      <sse:Security soapenv:mustUnderstand="1" xmlns:sse="http://example.com/security.xsd">
         <sse:UsernameToken wsu:Id="UsernameToken-9993341" xmlns:wsu="http://example.com/user.xsd">
            <sse:Username>user1</sse:Username>
            <sse:Password Type="http://example.com/password#PasswordText">1234</sse:Password>
         </sse:UsernameToken>
      </sse:Security>
   </soapenv:Header>
  <soapenv:Body>
    ....The rest of the transaction
  </soapenv:Body>
</soapenv:Envelope>

“mustUnderstand”属性没有 soapenv 前缀，sse:Security 标签立即关闭而不是将其他标签作为子标签，并且 UserName 的格式不正确

<sse:Username>user1</sse:Username>

.如何使用 SOAPElement 正确格式化消息方法？我需要知道的最重要的事情是如何正确地在安全标签内放置标签，以及如何正确格式化用户名/密码标签。

我已经尝试了 addHeaderElement 和 addChildElement 方法的不同组合，但我无法正确格式化它，而且 javadocs 没有提供关于它们将生成什么的足够详细信息。

最佳答案

这是从我的工作处理程序中获取的。希望它对你有用。

public static final String WSSE_NS = "http://docs.oasis-open.org/wss/2004/01/oasis-200401-wss-wssecurity-secext-1.0.xsd";
public static final String PASSWORD_TEXT_TYPE = "http://docs.oasis-open.org/wss/2004/01/oasis-200401-wss-username-token-profile-1.0#PasswordText";
public static final String WSSE_SECURITY_LNAME = "Security";
public static final String WSSE_NS_PREFIX = "wsse";

private String username;
private String password;
private boolean mustUnderstand = false;

public boolean handleMessage(SOAPMessageContext messageContext) {
    Object bOutbound = messageContext.get(MessageContext.MESSAGE_OUTBOUND_PROPERTY);
    if (bOutbound == Boolean.TRUE) {
        try {
            if (username != null && username.length() != 0) {
                addSecurityHeader(messageContext);
                LOG.debug("Added security header");
            } else {
                LOG.debug("No username configured thus not adding a security header");
            }
        } catch (Exception e) {
            LOG.error("Exception in handleMessage", e);
            return false;
        }
    }
    return true;
}

private void addSecurityHeader(SOAPMessageContext messageContext) throws SOAPException {
    SOAPFactory sf = SOAPFactory.newInstance();
    SOAPHeader header = messageContext.getMessage().getSOAPPart().getEnvelope().getHeader();
    if (header == null) {
        header = messageContext.getMessage().getSOAPPart().getEnvelope().addHeader();
    }

    Name securityName = sf.createName(WSSE_SECURITY_LNAME, WSSE_NS_PREFIX, WSSE_NS);
    SOAPHeaderElement securityElem = header.addHeaderElement(securityName);
    securityElem.setMustUnderstand(mustUnderstand);

    Name usernameTokenName = sf.createName("UsernameToken", WSSE_NS_PREFIX, WSSE_NS);
    SOAPElement usernameTokenMsgElem = sf.createElement(usernameTokenName);

    Name usernameName = sf.createName("Username", WSSE_NS_PREFIX, WSSE_NS);
    SOAPElement usernameMsgElem = sf.createElement(usernameName);
    usernameMsgElem.addTextNode(username);
    usernameTokenMsgElem.addChildElement(usernameMsgElem);

    Name passwordName = sf.createName("Type", WSSE_NS_PREFIX, WSSE_NS);
    SOAPElement passwordMsgElem = sf.createElement("Password", WSSE_NS_PREFIX, WSSE_NS);

    passwordMsgElem.addAttribute(passwordName, PASSWORD_TEXT_TYPE);
    passwordMsgElem.addTextNode(password);
    usernameTokenMsgElem.addChildElement(passwordMsgElem);

    securityElem.addChildElement(usernameTokenMsgElem);
}

关于java - 如何使用自定义 SOAPHandler 正确格式化 SOAP 消息信封，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/1197355/