RE: [xsl] XSL:FO Identify Transforms

[Chris Booth <Chris.Booth@xxxxxxxxxxxxxxxx>]

Thanks Ken,

Based on this template:
<xsl:template match="node() | @*">
	<xsl:copy>
		<xsl:apply-templates select="@* | node()" />
	</xsl:copy>
</xsl:template>

I was trying to construct in FO, but not quite there. I feel I am missing your
suggestion


				<fo:flow flow-name="xsl-region-body">
					<fo:table>
					<fo:table-column column-width="200mm"/>
						<fo:table-body border="1pt solid red">
							<fo:table-row text-align="center">
								<fo:table-cell>
									<xsl:for-each select="node() | @*">
										<fo:block>
											<xsl:for-each select="@* | node()">
												<fo:inline>
													<xsl:value-of select="." />
												</fo:inline>
											</xsl:for-each>
										</fo:block>
									</xsl:for-each>
								</fo:table-cell>
							</fo:table-row>
						</fo:table-body>
					</fo:table>
   				</fo:flow>
Reagrds

Chris
-----Original Message-----
From: G. Ken Holman [mailto:gkholman@xxxxxxxxxxxxxxxxxxxx]
Sent: 16 September 2011 17:33
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Subject: Re: [xsl] XSL:FO Identify Transforms

At 2011-09-16 17:19 +0100, you wrote:
>I have an XSL I use for identifying the XML - e.g. takes the XML and
>transforms it into another XML. This works as expected when it's XSLT.
>However I need to make this work in the same work but using XSL:FO so
>that it can be made pretty for print and outputted as a PDF.
>
>I am sure this should be simple, and I am missing something obvious.

A user's vocabulary cannot simply be copied into an XSL-FO file, just in the
same way it cannot simply be copied into an HTML file.

>How can I get the <xsl:copy> and <xsl:elements> working in FO, or
>something similar?

By not using those instructions and translating your input into appropriate
XSL-FO constructs.

For example, match on your block-level constructs and use <fo:block>, and
match on your inline-level constructs and use <fo:inline>.

><xsl:template match="/ | node()">
>                 <xsl:copy>
>                    <xsl:apply-templates select="@* | node()" />
>                 </xsl:copy>
></xsl:template>

Above you are copying your input

><xsl:template match="@*" >
>                 <xsl:element name="attribute">
>                                 <xsl:attribute name="name">
>                                                 <xsl:value-of
>select="local-name()" />
>                                 </xsl:attribute>
>                                 <xsl:value-of select="." />
>                 </xsl:element>
></xsl:template>

Unrelated to your question, the above can be replaced with:

<xsl:template match="@*"><xsl:copy/></xsl:template>

.... or even incorporated into a single template rule with:

<xsl:template match="node() | @*">
                 <xsl:copy>
                    <xsl:apply-templates select="@* | node()" />
                 </xsl:copy>
</xsl:template>

... but it is still inappropriate in your stylesheet to go to XSL-FO.

>SAMPLE XML:
><?xml version="1.0" encoding="UTF-8"?>
><xml>
>             <header>
>                         <test>This is  test</test>
>             </header>
></xml>

The XSL-FO specification tells an engine to tolerate elements it does not
recognize when those elements are not in the XSL-FO namespace, but it throws
them and their content away.

The HTML specification tells a browser to tolerate elements it does not
recognize, but it still processes the content of that unrecognized element.

Again, draw the parallel to HTML:  if you used the identity transform in a
stylesheet going to HTML, your end result would have <header> and <test>
elements in the HTML and you wouldn't get a pretty print because a browser
does not recognize your vocabulary.  In HTML you have to build a document of
<div> and <span> (or whatever HTML) from your XML.

Similarly, when using XSL-FO you have to build a document of <block> and
<inline> (or whatever XSL-FO) from your XML, because the engine does not
recognize your vocabulary.

I hope this helps.

. . . . . . . . . . .  Ken


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