How to get everything within an XML element with XSLT

Question

I'm trying to get XSLT to transform something like

<foo>
    lorem ipsum
    <br />
    some more lorem ipsum
</foo>

into

<p>
    lorem ipsum
    <br />
    some more lorem ipsum
</p>

I've tried countless paths for the select attribute of

<p><xsl:value-of select="/foo/*" /></p>

But none of them seemed to work. Is it even possible to get all the children of an element as plain text?

JLRishe · Accepted Answer · 2013-04-04 19:24:45Z

3

You can use xsl:copy-of:

<p><xsl:copy-of select="/foo/node()" /></p>

But a preferable approach would be to add an identity template to your XSLT:

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

And then make use of that:

<p><xsl:apply-templates select="/foo/node()" /></p>

answered Apr 4, 2013 at 19:24

JLRishe

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3 Comments

Simon Over a year ago

Could you clarify why the xsl:template method is preferable? IF both methods have the same result, isn't the shortest method prefered?

JLRishe Over a year ago

XSLT is a templating language, and the preferred way to use it is to use copy-ofs and procedural-style constructs, and rather use a push-style programming style, allowing the XSLT to be run through templates, which might pick up and operate on different parts. Suppose that somewhere down the road you wanted this XSLT to convert any <b>s in the XML to <strong>s. The apply-templates approach would allow you to do that easily, but the copy-of approach would not allow you to do it at all.

JLRishe Over a year ago

In this portion of my comment "the preferred way to use it is to use copy-ofs and procedural-style constructs", I meant to say "the preferred way to use it is to avoid copy-ofs and procedural-style constructs".

Stefan · Accepted Answer · 2013-04-04 19:48:48Z

0

Let me know if the following works for you:

<?xml version="1.0" encoding="ISO-8859-1"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="/">  
  <xsl:for-each select="item/foo">
    <p>
      <xsl:value-of select="current()"/>
    </p>    
  </xsl:for-each> 
</xsl:template>
</xsl:stylesheet>

Tested with this XML:

<?xml version="1.0" encoding="ISO-8859-1"?>
<item>
<foo>
    lorem ipsum
    <br />
    some more lorem ipsum
</foo>
<foo>
    lorem ipsum
    <br />
    some more lorem ipsum
</foo>
</item>

answered Apr 4, 2013 at 19:48

Stefan

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2 Comments

Simon Over a year ago

I guess that would work, but the tag <foo> won't be repeated in the XML file, so it seems kind of redundant to write an xsl:for-each for a single node.

JLRishe Over a year ago

This approach would eliminate the <br/> elements (and any other elements that might be within the foos.

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