Copy child nodes with exeptions in XSLT 1.0

Question

I think the main question is how write exeption expression to get rid of "S-3" and "Z" attributes in the output result. But how?

source code

<root>
  <B att-1="some value" att-2="value"> text 1 
    <S-1>text 2</S-1>
    <S-2>text 3</S-2>
    <S-3 trash-att="value"> trash-text a </S-3>

    <Z z-att="value"> z-text 1 </Z>  
 </B>

  <B att-1="some value" att-2="value"> text 4
      <S-1> text 5</S-1>
      <S-2> text 6</S-2>
      <S-3 trash-att="value"> trash-text b </S-3>

      <Z z-att="value"> z-text 2 </Z>  
  </B>

  <B att-1="some value" att-2="value"> text 7
    <S-1> text 8</S-1>
    <S-2> text 9</S-2>
    <S-3 trash-att="value"> trash-text c </S-3>

    <Z z-att="value"> z-text 3 </Z>  
 </B>

</root>

desired output

<root>
  <B att-1="some value"
      att-2="value"
      S-1="text 2"
      S-2="text 3">

    <Z attr=" z-text 1 "/>  
  </B>

  <B att-1="some value"
      att-2="value"
      S-1=" text 5"
      S-2=" text 6">

      <Z attr=" z-text 2 "/>  
  </B>

  <B att-1="some value"
      att-2="value"
      S-1=" text 8"
      S-2=" text 9">

    <Z attr=" z-text 3 "/>  
   </B>

</root>

my xslt code (S-3 and Z attributes are still in place, but shouldn't) https://xsltfiddle.liberty-development.net/3NSTbfj/1

<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
    xmlns:msxsl="urn:schemas-microsoft-com:xslt" exclude-result-prefixes="msxsl">
  <xsl:output method="xml" indent="yes"/>

  <xsl:template match="@* | node()">
    <xsl:copy>
      <xsl:apply-templates select="@* | node()"/>
    </xsl:copy>
  </xsl:template>


<!-- 1 - MAIN transform with element B -->
  <xsl:template match="B">
    <B>
     <xsl:copy-of select="@*"/>
         <xsl:for-each select="*" >
        <xsl:attribute name="{name()}">
          <xsl:value-of select="text()"/>
        </xsl:attribute>
      </xsl:for-each>

        <xsl:apply-templates select="@*|node()"/>
   </B> 
   </xsl:template>


 <!-- 2 - keep additional Z node -->  
   <xsl:template match="Z">
    <Z > 
  <xsl:attribute name="attr">
        <xsl:value-of select="."/>
      </xsl:attribute>
    </Z> 
   </xsl:template>


  <!-- 3 - delete nodes -->
  <xsl:template match="S-1">
    <xsl:apply-templates/>
  </xsl:template>

  <xsl:template match="S-2">
    <xsl:apply-templates/>
  </xsl:template>

  <xsl:template match="S-3">
    <xsl:apply-templates/>
  </xsl:template>

</xsl:stylesheet>

I will appreciate any solution

Answer 1

You Can Try This and use for-each

    <?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
    xmlns:xs="http://www.w3.org/2001/XMLSchema"
    exclude-result-prefixes="xs"
    version="2.0">
    <xsl:output method="xml" omit-xml-declaration="no" indent="yes"/>
    <xsl:template match="root">
        <xsl:copy>
            <xsl:apply-templates select="B"/>
        </xsl:copy>
    </xsl:template>
    <xsl:template match="B">
        <xsl:copy>
            <xsl:copy-of select="@*"/>
            <xsl:for-each select="*[not(@*)]">
                <xsl:attribute name="{local-name()}">
                    <xsl:value-of select="."/>
                </xsl:attribute>
            </xsl:for-each>
            <xsl:apply-templates select="Z"/>
        </xsl:copy>
    </xsl:template>
    <xsl:template match="Z">
        <xsl:copy>
                <xsl:attribute name="attr">
                    <xsl:value-of select="."/>
                </xsl:attribute>
        </xsl:copy>
    </xsl:template>
</xsl:stylesheet>

DEMO:https://xsltfiddle.liberty-development.net/3NSTbfj/2