Re: [Simple-support] XPath serializer.read

[Niall G. <gal...@ya...>]

Hi,

Well if they are ordered you can do.

@Default
public class Root {

@path("type1[1]")
private final Type1Class type;
}

However, the XPath expressions in the @Path annotation is limited. I will be improving this in a future release. However, right now, its got limited functionality.

Niall

--- On Thu, 24/3/11, Francois D. Barnard <fba...@gm...> wrote:

From: Francois D. Barnard <fba...@gm...>
Subject: [Simple-support] XPath serializer.read
To: sim...@li...
Received: Thursday, 24 March, 2011, 3:43 AM

hi
 
I have a xml structure looking something like this
 
[data.xml]
<root>
 <type1 id="1_1"/>
 <type1 id="1_2"/>
 <type1 id="1_3"/>
 
 <type2 id="2_1"/>
 <type2 id="2_2"/>
 <type2 id="2_3"/>

 
 <type3 id="3_1"/>
 <type3 id="3_2"/>
 <type3 id="3_3"/>
</root>
 
 
I created a class for root and a class each for the different types.
 
The root class has lists of type something like
 
List<Type1Class> Type1List
 
List<Type2Class> Type2List
 
List<Type3Class> Type3List
 
All this is working fine. The problem i have is this file isgetting bigger by the day and i am working on a android so resources are limited.
 
 
So the question is is there a way for me to serialize/deserialize this big xml file by providing xpath to determine what node i want as the source.
 
something like this
 
Type1Class  tmp = serializer.read(Type1Class.class,"data.xml","/root/type1[@id='1_2']");
 
 
thanks


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