Re: [xsl] Removing nodes with text content ?

[Torsten Schassan <schassan@xxxxxx>]

Hi Fabien,

as Martin's answer didn't let you get rid of the line breaks,

> I would like to remove the intermediary<phrase>  nodes to avoid line
> breaks (which are highly desirable on other places, but here are
> annoying me), and thus having everything on a single line (using the
> docbook stylesheets, already customized).

a very specific, thus "inelegant" full "answer" below, but some 
additional hints anyway:

- to avoid line breaks in the output using normalize-space() is often useful
- xsl:strip-space might help as well.


> <xsl:template match="phrase[@role='rtf']">
>     	<xsl:copy>
>     		<xsl:apply-templates select="./*"/>
>     	</xsl:copy>
> </xsl:template>

1. If you want to remove the element, you shouldn't copy it. (On the 
other hand, the template for para lacks this.)

2. Why wasn't text in the output? Your select selects only element 
nodes. In the more general template you use correctly "node()".

This should be working then (XSLT1):

    <xsl:output indent="yes"/>

    <xsl:template match="para">
        <xsl:copy>
            <phrase>
                <xsl:if test=".//phrase[@role='rtf']">
                    <xsl:attribute name="role">rtf</xsl:attribute>
                </xsl:if>
                <xsl:apply-templates/>
            </phrase>
        </xsl:copy>
    </xsl:template>
    <xsl:template match="phrase[@role='rtf']" >
        <xsl:apply-templates select="node()"/>
        <xsl:if test="following-sibling::phrase"><xsl:text> 
</xsl:text></xsl:if>
    </xsl:template>

    <xsl:template match="text()">
        <xsl:value-of select="normalize-space(.)"/>
    </xsl:template>

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


Best, Torsten

--
Torsten Schassan
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