Re: [xsl] merging sequences

[David Carlisle <davidc@xxxxxxxxx>]

the problem as specified seemed under constrained eg if teh lsists were 
"x" and "y" do you output x,y or y,x, in the following I prefer the 
first list so would output x,y.



$ saxon9 -it m merge.xsl
N112 N100 N107 P2010 N109 P2014 P2015 N108 N203 N306 N206 N307 N311


<xsl:stylesheet version="2.0"
		xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
		xmlns:f="data:,f"
		>

<xsl:output method="text"/>

<xsl:variable name="a" select="
'N112','N100','N107','P2010','N109','P2014','P2015','N108','N203','N206','N307','N311'
"/>

<xsl:variable name="b" select="
'N112','N100','P2014','P2015','N108','N203','N306','N206','N307','N311'
"/>

<xsl:template name="m">

<xsl:sequence select="f:merge($a,$b)"/>
</xsl:template>

<xsl:function name="f:merge">
<xsl:param name="a"/>
<xsl:param name="b"/>
<xsl:choose>
 <xsl:when test="empty($a)">
  <xsl:sequence select="$b"/>
 </xsl:when>
 <xsl:when test="empty($b)">
  <xsl:sequence select="$a"/>
 </xsl:when>
 <xsl:when test="$a[1]=$b[1]">
  <xsl:sequence 
select="$a[1],f:merge($a[position()!=1],$b[position()!=1])"/>
 </xsl:when>
 <xsl:when test="$a[1]=$b">
  <xsl:sequence select="f:merge($a[position()!=1],$b)"/>
 </xsl:when>
 <xsl:otherwise>
  <xsl:sequence select="$a[1],f:merge($a[position()!=1],$b)"/>

 </xsl:otherwise>
</xsl:choose>
</xsl:function>
</xsl:stylesheet>


David


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