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From "xueqiang.mi" <allo...@gmail.com>
Subject Re: [jira] Created: (CAMEL-1392) groovy renderer
Date Wed, 12 Aug 2009 08:04:44 GMT

Hi,
I want to understand how RouteContext or Registry register and then look up
a new Processor object when  receive some DSL which contains new Processor
object, like the following one:
{code}
MyValidator validator = new MyValidator();
from("direct:start")
                    .doTry()
                        .process(validator).to("mock:valid")
                    .doCatch(ValidationException.class)
                        .to("mock:invalid")
{code}
1. Does the route/processor definition maintain the processor name
"validator" somewhere? 
2. If the processor is registered, how can I get the object and translate it
into a processor reference when rendering it. I want to render the above DSL
into:
{code}
from("direct:start")
                    .doTry()
                        .processRef("someName or some
registeredKey").to("mock:valid")
                    .doCatch(ValidationException.class)
                        .to("mock:invalid")
{code}

Thanks

JIRA jira@apache.org wrote:
> 
> groovy renderer
> ---------------
> 
>                  Key: CAMEL-1392
>                  URL: https://issues.apache.org/activemq/browse/CAMEL-1392
>              Project: Apache Camel
>           Issue Type: Sub-task
>             Reporter: James Strachan
> 
> 
> 
> 
> -- 
> This message is automatically generated by JIRA.
> -
> You can reply to this email to add a comment to the issue online.
> 
> 
> 

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