Simple XSLT transformation

Question

I have the following input file:

<!-- input.xml --> 
<?xml version="1.0" encoding="UTF-8"?>
<input>
  <value>
    aaa
    <value>
      bbb
      <value>ccc</value>
    </value>
  </value>
</input>

Expected output:

<!-- output.xml --> 
<?xml version="1.0" encoding="UTF-8"?>
<ul>
  <li>aaa</li>
  <li>bbb</li>
  <li>ccc</li>
</ul>

How should the XSLT file look like? I tried the following:

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:template match="/input">
    <ul>
      <xsl:for-each select="//value">
        <li> <xsl:value-of select="."/></li>
      </xsl:for-each>
    </ul> 
  </xsl:template>
</xsl:stylesheet>

but the the first result (aaa) also contains values from its subtree (bbb, ccc).

Answer 1

Try select="text()" instead of select="." to only select the text content.

Answer 2

The key to success in your case is that the template matching value should replicate only text from the current level :

<xsl:value-of select="text()"/>

Actually, it is better to add normalize-space() to remove "additional" white-space chars:

<xsl:value-of select="normalize-space(text())"/>

After printing out of the content corresponding to the current level li element, you should put apply-templates for child nodes only:

<xsl:apply-templates select="*"/>

So the whole script can look like below:

<?xml version="1.0" encoding="UTF-8" ?>
<xsl:transform xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="2.0">
  <xsl:output method="xml" encoding="UTF-8" indent="yes" />
  <xsl:strip-space elements="*"/>

  <xsl:template match="input">
    <ul>
      <xsl:apply-templates/>
    </ul>
  </xsl:template>

  <xsl:template match="value">
    <li>
      <xsl:value-of select="normalize-space(text())"/>
    </li>
    <xsl:apply-templates select="*"/>
  </xsl:template>
</xsl:transform>

For a working example see http://xsltransform.net/pNvs5w2