Subj : Re: E4X: Filtering Predicate Operator
To   : McKinley, Richard
From : McKinley, Richard
Date : Mon Aug 23 2004 07:39 pm

This is strange.  Some principle must be lost on me. Using this:
var tree =  <tree>
         <node level="0">
           <node level="1">
             <node level="2"/>
             <node level="2"><node level="3"/></node>
             <node level="2"><node level="3"/></node>
           </node>
         </node>
       </tree>;
the node gets found.  Using this:
var tree =  <tree>
         <node level="0">
           <node level="1">
             <node level="2"><node level="3"/></node>
             <node level="2"><node level="3"/></node>
           </node>
         </node>
       </tree>;
the node does not get found.  Why would that be?

Thanks,

McKinley


McKinley, Richard wrote:

> I have tried this two different ways.  The following code matches the 
> examples in section 11.2.4 very closely.  The results I want are all 
> nodes with a level of 1:
>           <node level="1">
>             <node level="2"/>
>             <node level="2"/>
>           </node>
>           <node level="1">
>             <node level="2">
>               <node level="3"/>
>             </node>
>           </node>
> See the code below.  Why am I only getting the first node with level="1"?
> 
> 
> 
>   var tree =  <tree>
>         <node level="0">
>           <node level="1">
>             <node level="2"/>
>             <node level="2"/>
>           </node>
>           <node level="1">
>             <node level="2">
>               <node level="3"/>
>             </node>
>           </node>
>         </node>
>       </tree>;
>   var nodesWithFilteringPredicateOperator = tree..node.(@level == 1);
>   out.println(nodesWithFilteringPredicateOperator.toXMLString());
>   /* output:
>   <node level="1">
>     <node level="2"/>
>     <node level="2"/>
>   </node>
>   */
> 
>   var nodesWithForEach = new XMLList();
>   var i = 0;
>   for each(var n in tree..node) {
>     with(n) {
>       if(@level == 1) {
>         nodesWithForEach[i++] = n;
>       }
>     }
>   }
>   out.println(nodesWithForEach.toXMLString());
>   /*output:
>   <node level="1">
>     <node level="2"/>
>     <node level="2"/>
>   </node>
>   */
> 
> Thanks,
> 
> McKinley

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