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我在应用服务器（如 JBoss、WebSphere 和 Weblogic）中运行了 Axis2 Web 服务，直到现在我在请求中传递用户详细信息并在处理用户之前对用户进行身份验证。

下一步是我想将身份验证位委托给 Java EE 应用程序服务器，一旦通过身份验证，应用程序服务器应该传递我将用作执行请求的上下文的 UserPrinciple。

我不确定我是否正确地问了这个问题？我想我将 WebContainer 身份验证与 WS-Security 的东西混合在一起。

任何人都可以通过一些我可以将其作为启动指南引用的文档为我指明正确的方向。

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2 回答 2

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好的，我已经尝试了一个解决方案，它在一定程度上奏效了。这是详细信息；

使用以下 services.xml 创建了 TestService

<serviceGroup>
    <service name="TestWebService" scope="application" targetNamespace="http://TestServiceWs"> 
        <schema schemaNamespace="http://TestServiceWs"/>
        <description>Test POJO Axis2 AAR deployment</description>
        <parameter name="ServiceClass">test.TestServiceWS</parameter>
        <messageReceivers>
            <messageReceiver mep="http://www.w3.org/2004/08/wsdl/in-only"
                class="org.apache.axis2.rpc.receivers.RPCInOnlyMessageReceiver"/>
            <messageReceiver mep="http://www.w3.org/2004/08/wsdl/in-out"
                class="org.apache.axis2.rpc.receivers.RPCMessageReceiver"/>
        </messageReceivers>
    </service>

    <module ref="rampart" />

    <parameter name="InflowSecurity">
      <action>
        <items>UsernameToken Timestamp</items>
        <passwordCallbackClass>test.PWHandlerServer</passwordCallbackClass>
      </action>
    </parameter>
</serviceGroup>

已实现PWHandlerServer.java

package test;

import java.io.IOException;

import javax.security.auth.callback.Callback;
import javax.security.auth.callback.CallbackHandler;
import javax.security.auth.callback.UnsupportedCallbackException;

import org.apache.ws.security.WSPasswordCallback;

public class PWHandlerServer implements CallbackHandler {

    // the username and password we expect incoming WS calls to use
    private String user = "wsuser";
    private String pwd = "wspwd";

    public void handle (Callback[] callbacks) throws IOException, UnsupportedCallbackException {
        for (int i = 0; i < callbacks.length; i++) {
            if (callbacks[i] instanceof WSPasswordCallback) {
                WSPasswordCallback pc = (WSPasswordCallback) callbacks[i];
                System.out.println("identifier: "+pc.getIdentifer()+", usage: "+pc.getUsage());

                if (pc.getUsage() == WSPasswordCallback.USERNAME_TOKEN) {
                    // for passwords sent in digest mode we need to provide the password,
                    // because the original one can't be un-digested from the message

                    // we can throw either of the two Exception types if authentication fails
                    if (! user.equals(pc.getIdentifer()))
                        throw new IOException("unknown user: "+pc.getIdentifer());

                    // this will throw an exception if the passwords don't match
                    pc.setPassword(pwd);

                } else if (pc.getUsage() == WSPasswordCallback.USERNAME_TOKEN_UNKNOWN) {
                    // for passwords sent in cleartext mode we can compare passwords directly

                    if (! user.equals(pc.getIdentifer()))
                        throw new IOException("unknown user: "+pc.getIdentifer());

                    // we can throw either of the two Exception types if authentication fails
                    if (! pwd.equals(pc.getPassword()))
                        throw new IOException("password incorrect for user: "+pc.getIdentifer());
                }
                // If everything is OK then set the context and move on
                TestRequestContext ctx = new TestRequestContext(pc.getIdentifer());
                TestRequestContext.setRequestContext(ctx);

            } else {
                throw new UnsupportedCallbackException(callbacks[i], "Unrecognized Callback");
            }
        }
    }
}

已实现TestRequestContext

package test;


class TestRequestContext {
    private final static ThreadLocal<TestRequestContext> currentContext = new ThreadLocal<TestRequestContext>();

    public static void setRequestContext(TestRequestContext context) {
        currentContext.set(context);
    }

    public static TestRequestContext getRequestContext() {
        return currentContext.get();
    }

    public static void clearRequestContext() {
        currentContext.remove();
    }

    private String userPrincipal;

    public TestRequestContext(String principal) {
        this.userPrincipal = principal;
    }

    public String getUserPrincipal(){
        return this.userPrincipal;
    }
}

现在我应该能够在我的 Web 服务类中访问我的 TestRequestContext.getUserPricipal() 并将其传递给 DB 以进行事务和切换安全上下文。唯一的问题是当我尝试通过 SoapUI 访问服务时，它给了我以下异常；

    15:07:34,924 INFO  [STDOUT] identifier: wsuser, usage: 5
15:08:48,081 INFO  [STDOUT] [ERROR] WSDoAllReceiver: security processing failed (actions mismatch)
org.apache.axis2.AxisFault: WSDoAllReceiver: security processing failed (actions mismatch)
        at org.apache.rampart.handler.WSDoAllReceiver.processBasic(WSDoAllReceiver.java:344)
        at org.apache.rampart.handler.WSDoAllReceiver.processMessage(WSDoAllReceiver.java:86)
        at org.apache.rampart.handler.WSDoAllHandler.invoke(WSDoAllHandler.java:72)
        at org.apache.axis2.engine.Phase.invoke(Phase.java:318)
        at org.apache.axis2.engine.AxisEngine.invoke(AxisEngine.java:254)
        at org.apache.axis2.engine.AxisEngine.receive(AxisEngine.java:160)
        at org.apache.axis2.transport.http.HTTPTransportUtils.processHTTPPostRequest(HTTPTransportUtils.java:173)
        at org.apache.axis2.transport.http.AxisServlet.doPost(AxisServlet.java:144)
        at javax.servlet.http.HttpServlet.service(HttpServlet.java:710)
        at javax.servlet.http.HttpServlet.service(HttpServlet.java:803)
        at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:290)
        at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:206)
        at org.jboss.web.tomcat.filters.ReplyHeaderFilter.doFilter(ReplyHeaderFilter.java:96)
        at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:235)
        at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:206)
        at org.apache.catalina.core.StandardWrapperValve.invoke(StandardWrapperValve.java:230)
        at org.apache.catalina.core.StandardContextValve.invoke(StandardContextValve.java:175)
        at org.jboss.web.tomcat.security.SecurityAssociationValve.invoke(SecurityAssociationValve.java:182)
        at org.jboss.web.tomcat.security.JaccContextValve.invoke(JaccContextValve.java:84)
        at org.apache.catalina.core.StandardHostValve.invoke(StandardHostValve.java:127)
        at org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:102)
        at org.jboss.web.tomcat.service.jca.CachedConnectionValve.invoke(CachedConnectionValve.java:157)
        at org.apache.catalina.core.StandardEngineValve.invoke(StandardEngineValve.java:109)
        at org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:262)
        at org.apache.coyote.http11.Http11Processor.process(Http11Processor.java:844)
        at org.apache.coyote.http11.Http11Protocol$Http11ConnectionHandler.process(Http11Protocol.java:583)
        at org.apache.tomcat.util.net.JIoEndpoint$Worker.run(JIoEndpoint.java:446)
        at java.lang.Thread.run(Thread.java:619)

在 SoapUI 中，我只是使用“身份验证”选项卡来设置用户名和密码并调用？我需要在调用之前设置其他任何东西吗？

另外，使用 ThreadLocal 的方法是正确的，还是我也可以通过其他方式访问该原理？

谢谢。

于 2012-07-26T14:21:51.267 回答
0

您也可以获取以下格式的 SOAP 标头中传递的用户名。在授权之前，您可以在操作中访问它。您可能不需要 TestRequestContext 类来存储用户名。如果有人不同意，请告诉我。

MessageContext msgContext = MessageContext.getCurrentMessageContext();                   
// Getting HttpServletRequest from Message Context 
String username = (String)msgContext 
                           .getProperty(RampartMessageData.USERNAME);
于 2015-12-01T16:41:08.773 回答