Get attributes of a parent node XSLT

Question

I want to conditionally format URL's in Apache FOP, for this, I want to check if the property is of type HYPERLINK then apply conditional formatting and convert it into an URL.

Below is my XML

<properties>
    <property type="CUSTOM" id="150" key="localizedfield">
        <name>Localized Text</name>
        <value>Test</value>
    </property>
    <property type="CUSTOM" id="149" key="textareafield">
        <name>Textarea</name>
        <value>My longer default text.</value>
    </property>
    <property type="HYPERLINK" key="ASSET_LINK">
        <name>Asset Link</name>
        <value>Test=https://test.com</value>
    </property>
    <property type="CUSTOM" key="VALIDITY">
        <name>Asset Availability</name>
        <value>Available</value>
    </property>
</properties>

The XSL which I am using for transformation looks something like below

<xsl:template name="table-row">
    <xsl:for-each select="properties/property">
        <xsl:apply-templates select="."/>
    </xsl:for-each>
</xsl:template>

<xsl:template match="property">
    <fo:table-cell >
        <fo:block >
            <xsl:choose>
                <xsl:when test="<check if type is HYPERLINK>">
                    <!-- Format as hyperlink -->
                </xsl:when>
                <xsl:otherwise>
                    <!-- format as normal text -->
                </xsl:otherwise>
            </xsl:choose>
        </fo:block>
    </fo:table-cell>
</xsl:template>

in the xsl:when condition I only get the name and value, how can I get the complete property node here so I can check if the type attribute is HYPERLINK and then format accordingly?

Answer 1

I think you want

<xsl:when test="@type='HYPERLINK'">