You should NOT register the processor yourself, this is done by the
template bean thingy automatically
On Thu, Sep 8, 2022 at 1:39 PM ski n <raymondmees...@gmail.com> wrote:
> OK, sorry, I actually read the documentation on bindings and did use
> {{MyProcessor}} as the bean name.
>
> But it didn't work.
>
> I register my bean/processor as:
>
> registry.bind("MyProcessor", new MyProcessor(""));
>
> But when I run it, I got:
>
> Failed to resolve endpoint: bean://MyProcessor-1?method=process due to: No
> bean could be found in the registry for: MyProcessor-1
>
> Don't know where the "-1" comes from. However when I register it with that
>
> registry.bind("MyProcessor-1", new MyProcessor(""));
>
> Then the route works but not with the routetemplate parameter (because it's
> already iniaitized. So I tried it with
>
> registry.bind("MyProcessor-1", MyProcessor.class);
>
>
> Caused by: org.apache.camel.ResolveEndpointFailedException: Failed to
> resolve endpoint: bean://MyProcessor-1?method=process due to:
> java.lang.NoSuchMethodException:
>
> I also tried it with method parameters, instead of the constructor. I first
> tried to put an extra parameter on the process method, but the interface
> for Processor doesn't allow that. So I tried to use method overloading, but
> then it's unclear how to pass the exchange object as a parameter.
>
> Raymond
>
>
>
>
> On Thu, Sep 8, 2022 at 12:06 PM Claus Ibsen <claus.ib...@gmail.com> wrote:
>
> > Hi
> >
> > No you cannot - you mix standard Java with Camel "parsing" the model when
> > it calls the configure() method.
> > It would be the same in regular Camel route.
> > You basically do standard Java code with a new constructor and pass in a
> > string literal. Camel is not in use at that point.
> >
> > In your template bean example, then you need to use {{MyProcessor}} as
> the
> > bean name.
> > See the IMPORTANT note at:
> >
> >
> https://camel.apache.org/manual/route-template.html#_binding_beans_to_route_template
> >
> >
> >
> > On Thu, Sep 8, 2022 at 11:53 AM ski n <raymondmees...@gmail.com> wrote:
> >
> > > Hi,
> > >
> > > I have a routetemplate as follows:
> > >
> > > routeTemplate("processortemplate")
> > > .templateParameter("out")
> > > .from("direct:in")
> > > .process("MyProcessor")
> > > .to("{{out}}");
> > >
> > > This works.
> > >
> > > The processor is registered an the called by reference. Now I added a
> > > constructor argument to the processor and I tried to call it like this:
> > >
> > > routeTemplate("processortemplate")
> > > .templateParameter("processerParam")
> > > .templateParameter("out")
> > > .from("direct:in")
> > > .process(new MyProcessor("{{processorParam}}"))
> > > .to("{{out}}");
> > >
> > > This doesn't work, because the parameter of the processor is parsed
> > before
> > > the template parameter is parsed. Result is that the parameter
> > > {{myProcessorParam}} is passed as literal param.
> > >
> > > I tried to come with up something like this:
> > >
> > > routeTemplate("processortemplate")
> > > .templateParameter("processerparam")
> > > .templateParameter("out")
> > > .templateBean("MyProcessor")
> > > .typeClass("com.foo.MyProcessor")
> > > .property("processerparam", "{{processerparam}}")
> > > .end()
> > > .from("direct:in")
> > > .to("bean:MyProcessor?method=process")
> > > .to("{{out}}");
> > >
> > > But this didn't pass the parameter as well. As a workaround I set
> > > constructor argument as a header and then get the header within the
> > > processor.
> > >
> > > Question:
> > >
> > > Is there a way to pass constructor arguments to a processor in another
> > way?
> > > I couldn't find anything at the routeTemplate page on how to handle
> > > processors.
> > >
> > > Raymond
> > >
> >
> >
> > --
> > Claus Ibsen
> > -----------------
> > http://davsclaus.com @davsclaus
> > Camel in Action 2: https://www.manning.com/ibsen2
> >
>
--
Claus Ibsen
-----------------
http://davsclaus.com @davsclaus
Camel in Action 2: https://www.manning.com/ibsen2
