如何通过WSO2 ESB中的简单代理服务调用安全代理服务

时间:2015-02-19 07:13:52

标签: apache wso2 wso2esb

我在WSO2 ESB中提供了安全代理服务。我跟着这个tutorial。 现在我想通过使用另一个代理服务来调用此服务。但是当我从SoapUI发送请求到简单的代理服务时,我遇到了一个错误。我无法弄清楚的是如何将安全配置传递给代理服务。

注意:我使用的是WSO2 ESB 4.8.1。

安全代理服务代码:

<?xml version="1.0" encoding="UTF-8"?>
<proxy xmlns="http://ws.apache.org/ns/synapse"
       name="EchoProxy2"
       transports="https,http"
       statistics="disable"
       trace="disable"
       startOnLoad="true">
   <target>
      <inSequence>
         <log level="full" separator="InSequence Log"/>
      </inSequence>
      <outSequence>
         <log level="full" separator="Out Sequence Called"/>
         <send/>
      </outSequence>
      <endpoint>
         <address uri="http://wi1-Inspiron-3521:8280/services/echo"/>
      </endpoint>
   </target>
   <parameter name="disableREST">true</parameter>
   <parameter name="ScenarioID">scenario2</parameter>
   <enableSec/>
   <policy key="conf:/repository/axis2/service-groups/EchoProxy2/services/EchoProxy2/policies/SigOnly"/>
   <description/>
</proxy>

简单代理服务代码:

<?xml version="1.0" encoding="UTF-8"?>
<proxy xmlns="http://ws.apache.org/ns/synapse"
       name="CallEchoProxy"
       transports="https,http"
       statistics="disable"
       trace="disable"
       startOnLoad="true">
   <target>
      <inSequence>
         <log level="full" separator="*Calling Proxy In Sequence"/>
      </inSequence>
      <outSequence>
         <log level="full"
              separator="Calling Proxy Out Sequence"/>
      </outSequence>
      <endpoint>
         <address uri="http://wi1-Inspiron-3521:8280/services/EchoProxy2"/>
      </endpoint>
   </target>
   <description/>
</proxy>

日志错误:

[2015-02-19 11:22:48,718] ERROR - ServerWorker Error processing POST request for : /services/EchoProxy2
org.apache.axis2.AxisFault: Missing wsse:Security header in request
    at org.apache.rampart.handler.RampartReceiver.setFaultCodeAndThrowAxisFault(RampartReceiver.java:180)
    at org.apache.rampart.handler.RampartReceiver.invoke(RampartReceiver.java:99)
    at org.apache.axis2.engine.Phase.invokeHandler(Phase.java:340)
    at org.apache.axis2.engine.Phase.invoke(Phase.java:313)
    at org.apache.axis2.engine.AxisEngine.invoke(AxisEngine.java:261)
    at org.apache.axis2.engine.AxisEngine.receive(AxisEngine.java:167)
    at org.apache.synapse.transport.passthru.ServerWorker.processEntityEnclosingRequest(ServerWorker.java:411)
    at org.apache.synapse.transport.passthru.ServerWorker.run(ServerWorker.java:183)
    at org.apache.axis2.transport.base.threads.NativeWorkerPool$1.run(NativeWorkerPool.java:172)
    at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1145)
    at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:615)
    at java.lang.Thread.run(Thread.java:744)
Caused by: org.apache.rampart.RampartException: Missing wsse:Security header in request
    at org.apache.rampart.RampartEngine.process(RampartEngine.java:146)
    at org.apache.rampart.handler.RampartReceiver.invoke(RampartReceiver.java:92)
    ... 10 more

问题: 我无法理解如何将安全配置传递给我的简单代理服务。因此它可以与安全服务通信并获取响应。

1 个答案:

答案 0 :(得分:0)

您拥有安全代理,并且您还拥有不安全的代理。当您向非安全代理发送消息时,它必须调用安全代理。因此,您的未安全代理必须调用安全端点。在这里,安全端点是您的安全代理。要调用安全端点,必须将WS-Security策略附加到端点配置。您可以参考this博客,其中介绍了如何调用安全端点。如果您正确地按照步骤操作,我想您可以解决此问题。