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Home/ Questions/Q 8524025
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Editorial Team
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Editorial Team
Asked: June 11, 20262026-06-11T07:34:29+00:00 2026-06-11T07:34:29+00:00

I want to create an xml in the following format <parm-list> <param> <NAME>somename</NAME> <VALUE>somevalue</VALUE>

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I want to create an xml in the following format

<parm-list>
<param>
<NAME>somename</NAME>
<VALUE>somevalue</VALUE>
</param>
<param>
<NAME>somename</NAME>
<VALUE>somevalue</VALUE>
</param>
<param>
<NAME>somename</NAME>
<VALUE>somevalue</VALUE>
</param>
<param>
<NAME>somename</NAME>
<VALUE>somevalue</VALUE>
</param>
</param-list>

What can i do if i don’t want a <PARAM> field with specific <NAME> in it?

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  1. Editorial Team
    Editorial Team
    2026-06-11T07:34:31+00:00Added an answer on June 11, 2026 at 7:34 am

    Although I really don’t think that Xstream is really appropriate for the task, it does seem to be possible.

    You could implement your own converter similar to one of those. If you take a look at Converter interface, you’ll see that you can easily skip an element by not writing anything to HierarchicalStreamWriter.

    public class ParamConverter implements Converter{
    
        boolean canConvert(Class type){
            return Param.class.equals(type);
        }
    
        public void marshal(Object source, HierarchicalStreamWriter writer, MarshallingContext context){
            Param param = (Param)source;
            if (NAME_TO_SKIP.equals(param.getName()){
                return;
            }
            // delegate to ReflectionConverter or something else appropriate.
        }   
    }
    

    To register converter simply call xStram.registerConverter(new ParamConverter());.

    For more information please read this tutorial on converters.

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