How to access a root attribute from a deeper level in XSLT - xslt

Question

I'm calling a template:
<table>
<xsl:apply-templates select="data/pics/row"/>
</table>
The template is
<xsl:template match="row">
<tr>
<xsl:for-each select="td">
<td border="0">
<a href="{#referencddePage}">
<img src="{pic/#src}" width="{pic/#width}" height="{pic/#height}"/>
</a>
</td>
</xsl:for-each>
</tr>
</xsl:template>
My XML is:
<?xml version="1.0" encoding="iso-8859-8"?>
<?xml-stylesheet type="text/xsl" href="xslFiles\smallPageBuilder.xsl"?>
<data pageNo="3" referencePage="xxxxxxxxxxxxxxx.xml">
<pics>
<row no="0">
<td col="0">
<pic src="A.jpg" width="150" height="120"></pic>
</td>
</row>
</pics>
</data>
I want the line :a h r e f="{#referencddePage}" to get the input from
the root, :a h r e f= "{#referencddePage}"..., but I'm already in the <td level>.

Answer 1

I want the line :a h r e
f="{#referencddePage}" to get the
input from the root :a h r e f=
"{#referencddePage}"... but I'm
already in the <td level>
In case it is a rule that the #referencePage attribute is always an attribute of the top element, then it can always be accessed as:
/*/#referencePage
Therefore, in your code you'll have:
<a href="{/*/#referencePage}">
I would recommend not to use <xsl:for-each> and to use only and`. In this way the resulting XSLT code is more understandable and can be more easily modified in the future:
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output omit-xml-declaration="yes" indent="yes"/>
<xsl:strip-space elements="*"/>
<xsl:template match="row">
<tr>
<xsl:apply-templates/>
</tr>
</xsl:template>
<xsl:template match="td">
<td border="0">
<a href="{/*/#referencePage}">
<xsl:apply-templates/>
</a>
</td>
</xsl:template>
<xsl:template match="pic">
<img src="{#src}" width="{#width}" height="{#height}"/>
</xsl:template>
</xsl:stylesheet>
When this transformation is applied on the provided XML document,
<data pageNo="3" referencePage="xxxxxxxxxxxxxxx.xml">
<pics>
<row no="0">
<td col="0">
<pic src="A.jpg" width="150" height="120"></pic>
</td>
</row>
</pics>
</data>
the wanted output is produced:
<tr>
<td border="0">
<a href="xxxxxxxxxxxxxxx.xml">
<img src="A.jpg" width="150" height="120"/>
</a>
</td>
</tr>
See how each template is so very simple. Also, the code is further simplified.
Now, instead of:
<img src="{pic/#src}" width="{pic/#width}" height="{pic/#height}"/>
we have only:
<img src="{#src}" width="{#width}" height="{#height}"/>

Answer 2

Use an XPATH that "jumps" to the top of the document with a leading slash, then walk down the tree:
/data/#referencePage
Applying it to your stylesheet:
<xsl:template match="row">
<tr>
<xsl:for-each select="td">
<td border="0">
<a href="{/data/#referencePage}">
<img src="{pic/#src}" width="{pic/#width}" height="{pic/#height}"/>
</a>
</td>
</xsl:for-each>
</tr>
</xsl:template>