Hi Lothar,
One possible approach would be to have a processor that receives the
notification and creates dynamically a route based on the notification
header to poll the directory.
Here is some pseudo code:
from(jms)
.process(new Processor() {
@Override
public void process(Exchange exchange) throws
Exception {
exchange.getContext().addRoutes(new RouteBuilder() {
public void configure() {
from("file://" + header + ? + sendEmptyMessageWhenIdle=true)
.to(somewhere)
}
}
})
The route could send an empty message when all the files are processed,
allowing you to detect that and stop it as shown here
http://camel.apache.org/how-can-i-stop-a-route-from-a-route.html
HTH,
On Tue, Nov 26, 2013 at 10:39 PM, Lothar Werzinger <[email protected]>wrote:
> Hi,
>
> I would like to create a route that responds to a trigger (e.g. from a
> queue) and then scans a directory which path is given as a header attribute
> in the trigger message where I can then route the found files through a set
> of other components.
>
> However if I connect a file component to an input component it
> automatically becomes an output component, so the "direct approach"
> from(...).to(file://...).to(...) obviously does not work.
>
> How can I create such a route?
>
> Thanks in advance!
> --
>
> Lothar Werzinger
>
> Principal Architect
>
> Tradescape, Inc. - Enabling Efficient Digital Marketplaces
>
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>
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>
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>
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--
Bilgin Ibryam
Apache Camel & Apache OFBiz committer
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Twitter: @bibryam <https://twitter.com/bibryam>
Author of Instant Apache Camel Message Routing
http://www.amazon.com/dp/1783283475
