title: URI to Spring DSL How do I pass in REST Get Parameters
body: <p>An example URI for my rest service is as follows:</p>
  
  <p><a href="http://xx.xx.xxx.xx:8080/myservice/service/encode?encrypt=true&amp;payload=11/11/2013%207:59:15%20AM&amp;ttl=10h" rel="nofollow">http://xx.xx.xxx.xx:8080/myservice/service/encode?encrypt=true&amp;payload=11/11/2013%207:59:15%20AM&amp;ttl=10h</a></p>
  
  <p>The service route I am using on the node with my service on it is as follows.  This works fine.  The route should bridge whatever we give it through to the service.</p>
  
  <pre><code>&lt;route id="my-server"&gt;
    &lt;from uri="fabric-camel:myClusterId:jetty:http://xx.xx.xxx.xx:8484/myservice/service?matchOnUriPrefix=true" /&gt;
    &lt;to uri="jetty:http://xx.xx.xxx.xx:8080/myservice/service?bridgeEndpoint=true&amp;amp;throwExceptionOnFailure=false" /&gt;
  &lt;/route&gt;
  </code></pre>
  
  <p>My trouble is with the client route.   The time component is working fine but my efforts to append anything to the GET have failed.  If I can figure out how to do this I can create any number of client examples.</p>
  
  <pre><code>&lt;route id="fabric-client" errorHandlerRef="errorHandler"&gt;
    &lt;from uri="timer://foo?fixedRate=true&amp;amp;period=1000"/&gt;
    &lt;to uri="fabric-camel:myClusterId"/&gt;
  &lt;/route&gt;
  </code></pre>
  
  <p>Can someone give me a leg up by converting the URI example above to Sprint DSL?  I am hoping you can show me how to do it. From that I can figure out how to pass in variables.  </p>