At 2003-12-29 15:23 -0500, Norma Yeazell wrote:
I have 3 randlist in a row within a paragraph, the first should be at left
the next two indented diffently. Can anyone give me a clue how to get this
accomplished?
...
<step1 id="step1-6">
<title>step1 title</title>
<para>This is a step 1.
<randlist id="randlist-6">
<item>Be careful not to short out battery terminal. </item>
<item>Do not smoke or use open flame near batteries. </item>
<item>Batteries may explode from spark.</item>
</randlist>
<randlist id="randlist-7">
<item>Sub-unordered list items start with a bullet indented 7mm for the
left type limit.</item>
<item>The items within a sub-unordered list shall not be separated with a
blank line.</item>
</randlist>
<randlist id="randlist-8">
<item>Subsub-unordered list items start with a bullet indented 14mm for
the left type limit.</item>
<item>The items within a sub-unordered list shall not be separated with a
blank line.</item>
</randlist>
</para>
<step2 id="step2-2">
<title>This is a sample title for a step 2</title>
<para>This is a step 2.</para>
</step2>
</step1>
This turns out to be simpler than originally thought. Provided your
subordinate display relationship is based on a sibling source relationship,
you can take advantage of XSL-FO's support of expressions as follows:
<xsl:template match="randlist">
<list-block start-indent="{count(preceding-sibling::randlist)} * 7mm">
<xsl:apply-templates/>
</list-block>
</xsl:template>
<xsl:template match="item">
<list-item provisional-distance-between-starts="1cm">
<list-item-label end-indent="label-end()">
<block>1</block>
</list-item-label>
<list-item-body start-indent="body-start()">
<block><xsl:apply-templates/></block>
</list-item-body>
</list-item>
</xsl:template>
In the above you are indenting 7mm for each list after the first list by
counting the number of preceding sibling randlist elements and multiplying
that count by 7mm.
I hope this helps.
................................ Ken
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