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社区首页 >问答首页 >迁移到CXF3.2.7 ->如何解决SOAP处理过程中与密码相关的安全错误?

迁移到CXF3.2.7 ->如何解决SOAP处理过程中与密码相关的安全错误?
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Stack Overflow用户
提问于 2018-11-16 13:20:18
回答 2查看 1.5K关注 0票数 0

上下文:我正在尝试将一个CXF服务从CXF2.2.2迁移到CXF3.2.7

Problem:每当我试图通过像SOAP这样的软件针对我的the服务触发基于HTTP的请求时,就会进行迁移:

代码语言:javascript
复制
<SOAP-ENV:Header>
      <wsse:Security SOAP-ENV:mustUnderstand="1">
        <wsse:UsernameToken wsu:Id="">
        <wsse:Username>sampleUser</wsse:Username>
        <wsse:Password>12345</wsse:Password>
        <wsse:PartnerID>samplePartner</wsse:PartnerID>
</wsse:UsernameToken>
</wsse:Security>
    <wsa:To>http://localhost:8080/sampleWs</wsa:To>
    <wsa:Action>http://localhost:8080/sampleWs/sampleAction</wsa:Action>
      <wsa:From>
      <wsa:Address>http://schemas.xmlsoap.org/ws/2004/08/addressing/role/anonymous</wsa:Address>
</wsa:From>
</SOAP-ENV:Header>

我从CXF安全模块收到以下安全异常:

代码语言:javascript
复制
18:11:29,250 WARNING [org.apache.cxf.phase.PhaseInterceptorChain] (http-127.0.0.1:8080-1) Interceptor for {http://ws.sampleWs.varun/}SampleWebService has thrown exception, unwinding now: org.apache.cxf.binding.soap.SoapFault: A security error was encountered when verifying the message
        at org.apache.cxf.ws.security.wss4j.WSS4JUtils.createSoapFault(WSS4JUtils.java:234) [cxf-rt-ws-security-3.2.7.jar:3.2.7]
        at org.apache.cxf.ws.security.wss4j.WSS4JInInterceptor.handleMessageInternal(WSS4JInInterceptor.java:341) [cxf-rt-ws-security-3.2.7.jar:3.2.7]
        at org.apache.cxf.ws.security.wss4j.WSS4JInInterceptor.handleMessage(WSS4JInInterceptor.java:176) [cxf-rt-ws-security-3.2.7.jar:3.2.7]
        at org.apache.cxf.ws.security.wss4j.WSS4JInInterceptor.handleMessage(WSS4JInInterceptor.java:87) [cxf-rt-ws-security-3.2.7.jar:3.2.7]
        at org.apache.cxf.phase.PhaseInterceptorChain.doIntercept(PhaseInterceptorChain.java:308) [cxf-core-3.2.7.jar:3.2.7]
        at org.apache.cxf.transport.ChainInitiationObserver.onMessage(ChainInitiationObserver.java:121) [cxf-core-3.2.7.jar:3.2.7]
        .
        .
        .
Caused by: org.apache.wss4j.common.ext.WSSecurityException: BSP:R4201: Any PASSWORD MUST specify a Type attribute
        at org.apache.wss4j.common.bsp.BSPEnforcer.handleBSPRule(BSPEnforcer.java:57) [wss4j-ws-security-common-2.2.2.jar:2.2.2]
        at org.apache.wss4j.dom.message.token.UsernameToken.checkBSPCompliance(UsernameToken.java:834) [wss4j-ws-security-dom-2.2.2.jar:2.2.2]
        at org.apache.wss4j.dom.message.token.UsernameToken.<init>(UsernameToken.java:143) [wss4j-ws-security-dom-2.2.2.jar:2.2.2]
        at org.apache.wss4j.dom.processor.UsernameTokenProcessor.handleUsernameToken(UsernameTokenProcessor.java:137) [wss4j-ws-security-dom-2.2.2.jar:2.2.2]
        at org.apache.wss4j.dom.processor.UsernameTokenProcessor.handleToken(UsernameTokenProcessor.java:62) [wss4j-ws-security-dom-2.2.2.jar:2.2.2]
        at org.apache.wss4j.dom.engine.WSSecurityEngine.processSecurityHeader(WSSecurityEngine.java:340) [wss4j-ws-security-dom-2.2.2.jar:2.2.2]
        at org.apache.cxf.ws.security.wss4j.WSS4JInInterceptor.handleMessageInternal(WSS4JInInterceptor.java:285) [cxf-rt-ws-security-3.2.7.jar:3.2.7]
        ... 28 more

根本原因似乎是wsse:password标记中缺少属性。

我检查了安全WS规范:http://docs.oasis-open.org/wss/2004/01/oasis-200401-wss-wssecurity-secext-1.0.xsd

在那里,我无法找到wsse:password的强制类型属性,正如最新CXF所期望的那样。

备注:CXF版本2.2.2很好地处理了相同的请求

问题:是否有一种方法可以防止CXF删除上述请求并允许其通过。我环顾四周,却找不到任何答案。有什么建议吗?或者修改SOAP请求是唯一的解决方案?

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回答 2

Stack Overflow用户

回答已采纳

发布于 2018-11-20 04:19:11

上面的请求被删除,因为正如@GPI前面所述,CXF试图强制执行基本安全规范(http://www.ws-i.org/profiles/basicsecurityprofile-1.1.html)的http://www.ws-i.org/profiles/basicsecurityprofile-1.1.html

为了防止CXF强制执行,我们可以使用ws-securitywss4j提供的配置常量来指示CXF停止强制执行。

解决方案#1 (使用WSS4J拦截器)

在spring配置文件中,可以将isBspCompliant设置为false

代码语言:javascript
复制
    <jaxws:inInterceptors>
        <bean class="org.apache.cxf.ws.security.wss4j.WSS4JInInterceptor">
            <constructor-arg>
                <map>
                    <entry key="action" value="UsernameToken" />
                    <entry key="passwordType" value="PasswordText" />
                    <entry key="passwordCallbackRef">
                        <ref bean="passwordCallback" />
                    </entry>
                    <entry key="isBSPCompliant" value="false"/>
                </map>
            </constructor-arg>
        </bean>
    </jaxws:inInterceptors>

解决方案#2 (使用JAX)

在spring配置文件中,可以将ws-security.is-bsp-compliant设置为false

代码语言:javascript
复制
    <jaxws:endpoint id="sample" implementor="sample.ws.SampleWebService" address="/SampleWebService">
    <jaxws:properties> 
    <entry key="ws-security.is-bsp-compliant" value="false"/> 
    </jaxws:properties> 
</jaxws:endpoint>

有关其他配置选项,可以参考以下页面:

http://ws.apache.org/wss4j/config.html

http://cxf.apache.org/docs/ws-securitypolicy.html

票数 2
EN

Stack Overflow用户

发布于 2020-08-06 16:21:26

我也面临着同样的问题,发现soap请求中存在一个问题。

我能够通过添加以下内容来解决这个问题:

  • soap中的以下属性:信封xmlns:wsse="http://docs.oasis-open.org/wss/2004/01/oasis-200401-wss-wssecurity-secext-1.0.xsd"
  • wsse中的以下属性:密码Type="http://docs.oasis-open.org/wss/2004/01/oasis-200401-wss-username-token-profile-1.0#PasswordText"

我请求的标题如下所示

代码语言:javascript
复制
<soapenv:Envelope xmlns:soapenv="http://schemas.xmlsoap.org/soap/envelope/" xmlns:tic="http://ticket.degroupage.atos.ma/"  xmlns:wsse="http://docs.oasis-open.org/wss/2004/01/oasis-200401-wss-wssecurity-secext-1.0.xsd">
   <soapenv:Header>
      <wsse:Security soapenv:mustUnderstand="1">
         <wsse:UsernameToken><wsse:Username>XXXX</wsse:Username>
         <wsse:Password Type="http://docs.oasis-open.org/wss/2004/01/oasis-200401-wss-username-token-profile-1.0#PasswordText">XXXX</wsse:Password>
         </wsse:UsernameToken>
     </wsse:Security></soapenv:Header>
   <soapenv:Body>
...
票数 0
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页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/53338727

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