How do I generate a XSLT stylesheet for the following TEI snipet?

Question

i wanted to ask you for help. I am totally new to XSLT and i wanted to know if someone can show me the right XSLT stylesheet for the following TEI-snipet:

<div>
    <head>Weitere Aufzählungen</head>
    <list rend="numbered">
        <item n="1">A</item>
        <item n="2">B</item>                           
        <item n="3">C<list rend="numbered">
            <item n="3.1">a</item>
            <item n="3.2">b</item>
            </list>
        </item>
    </list>
</div>

The Output should look like that in the HTML-Document:

1. A
2. B
3. C
    3.1 a
    3.2 b

Thank you so much for helping me :)

Well, which HTML markup do you want to generate exactly? The exact look is not something XSLT creates. And have you tried to start reading an XSLT book like https://cranesoftwrights.github.io/books/ptux/index.htm or an online tutorial? — Martin Honnen, Oct 27 '18 at 15:04
Do those `item/@n` attributes always start with the index `1`, as in the example (e.g. `1,2,3`, `3.1,3.2`)? Or can there be lists where you want the numbering to be started with a different number? — Martin Honnen, Oct 27 '18 at 16:22
No, ist should be outlayed as above, but instead of 1,2,3, it can be also I,II,III and than, you know III.1, III.2 and so on. That does not matter. — CHRISTOPH HARTLEB, Oct 27 '18 at 16:29

score 0 · Accepted Answer · answered Oct 27 '18 at 15:19

0

If you want a text output, the following stylesheet with <xsl:output method="text" /> will do. It distinguishes the level of indentation by counting the item ancestor nodes and adds an extra . on level 0.

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:output method="text" indent="yes" />
  <xsl:variable name="newLine"  select="'&#xa;'" />

    <xsl:template match="text()" />

    <xsl:template match="/div/list">
        <xsl:apply-templates select="item" />
    </xsl:template>

    <xsl:template match="item">
        <xsl:for-each select="ancestor::item"><xsl:text>   </xsl:text></xsl:for-each>
        <xsl:value-of select="@n" />
        <xsl:if test="not(ancestor::item)"><xsl:text>.</xsl:text></xsl:if>
        <xsl:value-of select="concat(' ',text(),$newLine)" />
        <xsl:apply-templates select="list" />
    </xsl:template>   
</xsl:stylesheet>

Output is:

1. A
2. B
3. C
   3.1 a
   3.2 b

BTW you may need to add an appropriate namespace declaration for the TEI namespace to the xsl:stylesheet element.

answered Oct 27 '18 at 15:19

zx485

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It didn't work out. I just added your code into the XSLT Sheet, but it did not appear in the HTML-Document – CHRISTOPH HARTLEB Oct 27 '18 at 16:13
If I use the tag, the whole document appears in plain text. But I want to have the output styled as it is in the normal document. Do you know what I can change here? Maybe I could send you the documents? – CHRISTOPH HARTLEB Oct 27 '18 at 16:16
You could set the output `method="html"`. But the result will still be text and no HTML `list` like `
– zx485 Oct 27 '18 at 18:01
Hey, sorry for very late reply, it worked out for me, thank you somuch for your help ;) – CHRISTOPH HARTLEB Dec 02 '18 at 18:43
Thanks for your appropriate reply and welcome to StackOverflow. – zx485 Dec 02 '18 at 21:23

How do I generate a XSLT stylesheet for the following TEI snipet?

1 Answers1