Reputation: 4350
I have an xml element:
<myElement>item1 item2 item3</myElement>
I want to use XSLT to transform it to:
<newElement>item1</newElement>
<newElement>item2</newElement>
<newElement>item3</newElement>
How woudl the xslt look? I am more interested in how to loop though the list of myElement or how to get it to become a list or variable.
Please advise.
Upvotes: 1
Views: 1652
Reputation: 163458
In XSLT 2.0 it's easier:
<xsl:stylesheet version="2.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output omit-xml-declaration="yes" indent="yes"/>
<xsl:template match="myElement">
<xsl:for-each select="tokenize(., '\s+')">
<newElement><xsl:value-of select="."/></newElement>
</xsl:for-each>
</xsl:template>
</xsl:stylesheet>
Upvotes: 1
Reputation: 243529
This transformation:
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output omit-xml-declaration="yes" indent="yes"/>
<xsl:template match="text()[contains(., ' ')]" name="tokenize">
<xsl:param name="pText" select="."/>
<xsl:if test="string-length($pText)">
<newElement>
<xsl:value-of select=
"substring-before(concat($pText,' '),' ')"/>
</newElement>
<xsl:call-template name="tokenize">
<xsl:with-param name="pText" select=
"substring-after($pText,' ')"/>
</xsl:call-template>
</xsl:if>
</xsl:template>
</xsl:stylesheet>
when applied on the provided XML document:
<myElement>item1 item2 item3</myElement>
produces the wanted, correct result:
<newElement>item1</newElement>
<newElement>item2</newElement>
<newElement>item3</newElement>
Upvotes: 2