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Subject:
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How to resolve IDREF in XSL
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Posted By:
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teahouse
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Post Date:
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9/15/2006 11:55:52 AM
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Hi,
I am a newbie to XML/XSLT. I searched on this forum, however did not
find exactly what I look up. Basically, I have two tables A and B,
A contains a REF (a foreign key to B), so when I display table A
using XSL, since I have the idref to B's entry, how can I extract
attribute from B?
For illustration purpose, I have a database DTD for tools and
customers, and sample tool.xml, and its corresponding tool.xsl.
How do I display the customer's name, instead of the current
refID (I'm using borrower/@bid) (which is cust1)? See all the file
below.
Thanks for help!
-Tao
================== tool.dtd ===============
<!ELEMENT database (tools|customers)*>
<!ELEMENT tools (tool)* >
<!ELEMENT tool (toolname, borrower*)>
<!ELEMENT toolname (#PCDATA)>
<!ELEMENT borrower EMPTY>
<!ATTLIST borrower bid IDREF #REQUIRED>
<!ELEMENT customers (customer)*>
<!ELEMENT customer (name, address) >
<!ATTLIST customer cid ID #REQUIRED>
<!ELEMENT name (#PCDATA)>
<!ELEMENT address (#PCDATA)>
================== tool.xml===============
<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE database SYSTEM "tool.dtd">
<?xml-stylesheet type="text/xsl" href="tool.xsl"?>
<database>
<tools>
<tool>
<toolname>Foo Bar</toolname>
<borrower bid="cust1" />
</tool>
</tools>
<customers>
<customer cid="cust1">
<name>Ben Jackson</name>
<address>688 Adams Alley, Cincinnati, OH</address>
</customer>
</customers>
</database>
================== tool.xsl ===============
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="html" omit-xml-declaration="no" doctype-system="http://www.w3c.org/TR/xhtml/DTD/xhtml1-strict.dtd" doctype-public="-//W3C//DTD XHTML 1.0 Strict//EN"/>
<xsl:template match="/"> <!-- match root element -->
<html lang="en" xml:lang="en" xmlns="http://www.w3.org/1999/xhtml">
<head>
<title>This is a Tool Collection</title>
</head>
<body>
<h2>My Tool Collection</h2>
<table border="1">
<tr bgcolor="#9acd32">
<th align="left">Tool Name</th>
<th>Borrower ID</th>
<th>Borrower Name</th>
</tr>
<xsl:for-each select="database/tools/tool">
<tr>
<td>
<xsl:value-of select="toolname"/>
</td>
<td>
<xsl:value-of select="borrower/@bid"/>
</td>
</tr>
</xsl:for-each>
</table>
</body>
</html>
</xsl:template>
</xsl:stylesheet>
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Reply By:
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joefawcett
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Reply Date:
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9/15/2006 12:25:45 PM
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There are at least three approaches, the first is to use something like:<xsl:for-each select="database/tools/tool">
<tr>
<td>
<xsl:value-of select="toolname"/>
</td>
<td>
<xsl:value-of select="borrower/@bid"/>
</td>
<td>
<xsl:value-of select="/*/customers/customer[@cid = current()/borrower/@bid]/name"/>
</td>
</tr>
</xsl:for-each>
The current() function refers to the context node before the XPath expression is evaluated, alternatively you could store the value in a variable.
The better way to do this is to declare a key to access the customers. <xsl:key name="customerById" match="customer" use="@cid"/>
This needs to be outside of any xsl:template element.
You can then use this to retrieve customers:<xsl:for-each select="database/tools/tool">
<tr>
<td>
<xsl:value-of select="toolname"/>
</td>
<td>
<xsl:value-of select="borrower/@bid"/>
</td>
<td>
<xsl:value-of select="key('customerById', borrower/@bid)/name"/>
</td>
</tr>
</xsl:for-each>
Thirdly, as you have defined @cid to be of type ID then you can use the id function:<xsl:for-each select="database/tools/tool">
<tr>
<td>
<xsl:value-of select="toolname"/>
</td>
<td>
<xsl:value-of select="borrower/@bid"/>
</td>
<td>
<xsl:value-of select="id(borrower/@bid)/name"/>
</td>
</tr>
</xsl:for-each>
--
Joe (Microsoft MVP - XML)
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Reply By:
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mhkay
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Reply Date:
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9/15/2006 12:35:37 PM
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You can use the id() function. If @ref is an IDREF attribute, then id(@ref) returns the element whose ID it contains.
This is rather dependent on ID's being notified to the XSLT processor. Sometimes this breaks down, for example if you add a preprocessing phase to your transformation pipeline. Some people therefore recommend using keys instead. Declare
<xsl:key name="customer-id" match="customer" use="id"/>
and then key('customer-id', @ref)
will give you the customer whose ID value is in @ref.
Michael Kay
http://www.saxonica.com/
Author, XSLT Programmer's Reference and XPath 2.0 Programmer's Reference
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Reply By:
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teahouse
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Reply Date:
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9/15/2006 1:14:42 PM
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Joe/Michael,
Thank you so much for the prompt replies.
I'll try them out.
Thanks again!!!
-Tao
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