Remove a leaf node based on attribute value

Question

I have this leaf structure and I Want to remove leafs that have attribute:d value="true".

<outer_element>
     <leaf name="abc">
                  <attribute:a value="1"/>
                  <attribute:b value="2"/>
                  <attribute:c value="3"/>
                  <attribute:d value="true"/>
         </leaf>


     <leaf name="xyz">
      ....
     </leaf>
</outer_element>

Here is the case I have written,

if(string(node-name($node)) = "leaf" )
then
let $flag :=
     for $child in $node/node()
 
         if(name($child) eq "d" and $child/@value eq "true") 
         then 
           return "true"
         else
           return "false"
             
 if (contains ($flag,"true"))
 then 
   ()
 else
   element
   {
     node-name($node)
   }
   {
     $node/@*
     ,
     for $child in $node/node()
         return my:filterfun($child)
   }

  else
    element
        {
        node-name($node)
        }           {
         $node/@*
       ,
       for $child in $node/node()
          return my:filterfun($child )
      }

I am having trouble putting up it right,getting errors. Any help is appreciated. Thanks

Which version of XQuery, which XQuery processor? Is XQuery update supported? — Martin Honnen, Jun 03 '22 at 09:01
Where do element names like `` come from? Is there a namespace declaration `xmlns:attribute="..."` on some ancestor elements you have not shown? — Martin Honnen, Jun 03 '22 at 09:54

score 0 · Answer 1 · answered Jun 03 '22 at 09:56

0

It seems like

outer_element ! <outer_element>
{
   * except leaf[attribute:d/@value = 'true']
}
</outer_element>

should do.

answered Jun 03 '22 at 09:56

Martin Honnen

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Remove a leaf node based on attribute value

1 Answers1