Re: [basex-talk] How to get list as result instead of a String

Hi Elango,

changing the serialization parameters, here: output:separator option, [1] could do the trick.

declare option output:separator "\n";

let $s := <interfaces> <interface> <name>intf0</name> </interface> <interface> <name>intf1</name> </interface> <interface> <name>intf2</name> </interface> <interface> <name>intf3</name> </interface> </interfaces> return $s/interface/name/text()

==>

intf0 intf1 intf2 intf3

Hope this helps, Alex

[1] http://docs.basex.org/wiki/Serialization

On 04.04.2013, at 11:47, elangovan MuthuSwamy elangovan.h@gmail.com wrote:

Hi All,

I am using XQuery to evaluate an xpath. My code is as follows,

/*Example xml I am using

<interface> <name>intf0</name> </interface> <interface> <name>intf1</name> </interface> <interface> <name>intf2</name> </interface> <interface> <name>intf3</name> </interface> */ XQuery query = new XQuery("/interfaces/interface/name/text()"); String str = query.execute(ctx);

// Result string is - intf0intf1intf2intf3

I want the result such as intf0 intf1 intf2 intf3

I could not get the result list and I could not differentiate whether the result has multiple values or a single value.

Please advice me if there is any other way to identify the result list.

Thanks in Advance.

-- Regards, Elango. _______________________________________________ BaseX-Talk mailing list BaseX-Talk@mailman.uni-konstanz.de https://mailman.uni-konstanz.de/mailman/listinfo/basex-talk