CODEWITHSUNDEEP
xmlxsltattributestransfer
MGL
MGL

Reputation: 47

Tranfer attribute into a new element

First of all I hope I will be understood.

I have this:

       <item>
             <ptr target="X"/>BlahBlah
       </item>

And I would like to convert it into this:

    <li>
        <a href="X">Blahblah</a>
    </li>

All I could do was create this:

<xsl:template match="tei:ptr">
        <li>
            <a>
                <xsl:value-of select="parent::node()"/>
            </a>
        </li>
        <xsl:apply-templates/>
    </xsl:template>

But the result wasn't the one I was waiting for:

<li>
      <a>BlahBlah</a>
</li>BlahBlah

I could change the elements I wanted but the content of the <item> element was displayed twice, and I ignore the way to display the href attribute. If required I can show my entire XSL sheet.

I searched through the stackoverflow without result, maybe I just don't know how to put my problem into words.

Could someone help and explain how it does work? I know I have little understanding of XSLT, but I'm trying.

Thank you very much for your answer,

Matthias

Upvotes: 0

Views: 36

Answers (1)

zx485
zx485

Reputation: 29022

One way to achieve this is:

<xsl:template match="//item/text()">
    <xsl:if test="normalize-space(.) != ''">
        <li>
            <xsl:element name="a">
                <xsl:attribute name="href">
                    <xsl:value-of select="../ptr/@target" />
                </xsl:attribute>
                <xsl:value-of select="normalize-space(.)" />
            </xsl:element>
        </li>
    </xsl:if>
</xsl:template>

which results in 

<?xml version="1.0" encoding="UTF-8"?>
<li>
    <a href="X">BlahBlah</a>
</li>

You might replace the //item with a relative path to the item element if appropriate.

Upvotes: 1

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