Here is what you can do .
from("direct:start")
.to("http://localhost/Files.xml";)
.split(xpath("//file/text()")).parallelProcessing()
.process(new Processor() {
@Override
public void process(Exchange
exchange) throws Exception {
String fileName =
exchange.getIn().getBody(String.class);
// Invoke copy
functionality based on the protocol [ local or remote ]
}
});
Where Files.xml looks like this..
<files>
<file>D:\test1.xml</file>
<file>D:\test3.xml</file>
<file>D:\test4.xml</file>
</files>
- Ravi
-----Original Message-----
From: sb.append [mailto:[email protected]]
Sent: Monday, March 24, 2014 6:51 PM
To: [email protected]
Subject: download files for each xml entry
Hello,
I've simple task to parse xml entries containing links to files. For each entry
the file should be downloaded. Is the following route a correct approach?
from("direct:start")
.to("http://address.of.xml";)
.filter()
.xpath("entry/link")
.to(body())
.to("file:input);
I'd like to add parallel download later too.
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