download files for each xml entry

["sb.append" <a2...@drdrb.net>]

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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RE: download files for each xml entry

[Ra...@cognizant.com]

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:a261427@drdrb.net]
Sent: Monday, March 24, 2014 6:51 PM
To: users@camel.apache.org
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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Re: download files for each xml entry

[Richard Kettelerij <ri...@gmail.com>]

Hi,

I don't think using a filter yields the desired outcome. With this filter
you're only verifying whether the XML file contains a "entry/filter" node.
If the XML file is valid this will always succeed but it will not help you
in downloading additional files.

I guess what you're looking for is the *splitter*. With the splitter you
can process each "entry/link" node separately. Read about it:
http://camel.apache.org/splitter.html. Also the splitter allows for easily
adding parallel processing latter on:
split(xpath("entry/link")).parallelProcessing()......

Regards,
Richard
http://richardlog.com


On Mon, Mar 24, 2014 at 2:21 PM, sb.append <a2...@drdrb.net> wrote:

> 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.
>
>
>
>
>
> --
> View this message in context:
> http://camel.465427.n5.nabble.com/download-files-for-each-xml-entry-tp5749276.html
> Sent from the Camel - Users mailing list archive at Nabble.com.
>