How to copy particular element in an XML using XSLT 1.0 -

first, let me provide sample xml guys clear after.

<a>1</a> <b>1</b> <c>1</c> <d>1</d> <e>1</e> <f>1</f> 

is possible copy node b , e f. need neglect node c , d.

there <xsl:copy> can copy elements, need particular element out of original xml.

thank you.

sure can remove needed elements. write empty templates on specified elements after identity transform.

source xml

<root>    <a>1</a>    <b>1</b>    <c>1</c>    <d>1</d>    <e>1</e>    <f>1</f> </root> 

xslt 1.0

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/xsl/transform"> <xsl:strip-space elements="*"/>    <!-- identity transform -->       <xsl:template match="@*|node()">     <xsl:copy>       <xsl:apply-templates select="@*|node()"/>     </xsl:copy>   </xsl:template>    <!-- empty template remove elements -->      <xsl:template match="c|d"/>  </xsl:stylesheet> 

output xml

<root>    <a>1</a>    <b>1</b>    <e>1</e>    <f>1</f> </root> 

alternatively, select particular nodes keep:

xslt 1.0

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/xsl/transform"> <xsl:strip-space elements="*"/>    <!-- root template match -->       <xsl:template match="root">     <xsl:copy>       <xsl:apply-templates select="a|b|e|f"/>     </xsl:copy>   </xsl:template>    <!-- select particular elements -->      <xsl:template match="a|b|e|f">       <xsl:copy-of select="."/>   </xsl:template>  </xsl:stylesheet>