problem transforming XML using xslt

micky3248 · August 10th, 2006, 08:32 AM

Hi,

I'm experiencing problems with XSL transformation. I have an XML file that looks like this:

Code:

<ZMSRubrik id="e95">
    <title><lang id="ger"><![CDATA[Here a title]]></lang></title>
    <ZMSTextarea id="e74"><text><lang id="ger"><![CDATA[Here a text]]></lang></text></ZMSTextarea>
    <ZMSRubrik id="e95">
        <title><lang id="ger"><![CDATA[Here Titel2]]></lang></title>
        <ZMSTextarea id="e74"><text><lang id="ger"><![CDATA[Text2]]></lang></text></ZMSTextarea>
    </ZMSRubrik>
</ZMSRubrik>

I want to transform this for FOP. So I use this XSL file:

Code:

    <fo:root>
        <fo:layout-master-set>
            <fo:simple-page-master master-name="A4">
                             ......
            </fo:simple-page-master>
        </fo:layout-master-set>
                <xsl:apply-templates />
    </fo:root>
<xsl:template match="ZMS|ZMSRubrik|ZMSDocument">
        <fo:page-sequence master-reference="A4">
            <fo:flow flow-name="xsl-region-body">
                <xsl:for-each select="title/lang">
                    <fo:block>
                        <xsl:value-of select="." />
                    </fo:block>
                </xsl:for-each>
                <xsl:apply-templates select="*[ZMSTextArea|ZMSGraphic]"/> 
            </fo:flow>
        </fo:page-sequence>

        <xsl:apply-templates select="*[ZMSRubrik|ZMSDocument]"/>
</xsl:template>
<xsl:template match="ZMSTextarea">
   here the transformation for textarea
</xsl:template>
<xsl:template match="ZMSGraphic">
   here the transformation for graphics
</xsl:template>

Using this, I thought I should get one page-sequence for each ZMSRubrik, and each page-sequence shouldn't be a child of another page-template. Unfortunately it doesn't seem to work. The 2 other templates (ZMSTextarea and ZMSGraphic) are for elements of a page and they work well. I really have no idea why my code doesn't work.

Thx for any help. Micky

mhkay · August 10th, 2006, 12:36 PM

I suspect you are under the impression that

select="*[ZMSTextArea|ZMSGraphic]"

will process children that are named ZMSTextArea or ZMSGraphic.

In fact it will process all children that have a child named ZMSTextArea or ZMSGraphic.

Perhaps you want

select="ZMSTextArea | ZMSGraphic"

Michael Kay
http://www.saxonica.com/
Author, XSLT Programmer's Reference and XPath 2.0 Programmer's Reference

micky3248 · August 10th, 2006, 01:05 PM

Oh, I think I misunderstood it. I just begun 2 weeks ago, so I still don't understand everything. I'll try it next week after my holiday... Thx Micky

micky3248 · August 18th, 2006, 02:40 AM

Hi Thx a lot, it works the way I want it. Unfortunately I'm still having problems. I wanted to use such a code:

Code:

<xsl:template match="ZMSTextarea">


    <xsl:variable name="format">
        <xsl:choose>
            <xsl:when test="format='headline_1'">headline_1</xsl:when>
            <xsl:when test="format='headline_2'">headline_2</xsl:when>
            <xsl:when test="format='headline_3'">headline_3</xsl:when>
            <xsl:when test="format='headline_4'">headline_4</xsl:when>
            <xsl:when test="format='headline_5'">headline_5</xsl:when>
            <xsl:when test="format='headline_6'">headline_6</xsl:when>
            <xsl:when test="format='indented_block'">indented_block</xsl:when>
            <xsl:when test="format='emphasis'">emphasis</xsl:when>
            <xsl:when test="format='caption'">caption</xsl:when>
            <xsl:when test="format='unordered_list'">unordered_list</xsl:when>
            <xsl:otherwise>standard</xsl:otherwise>
        </xsl:choose>
    </xsl:variable>

    <xsl:element name="fo:block" use-attribute-sets="{$format}">
        <xsl:for-each select="text/lang">
            <xsl:value-of select="." disable-output-escaping="yes"/>
        </xsl:for-each>
    </xsl:element>

</xsl:template>

but saxon is giving an error: "No attribute-set exists named {$format}"
Isn't it possible to use a variable instead of a fix name for use-attribute-sets?

mhkay · August 18th, 2006, 02:49 AM

Isn't it possible tu use a variable instead of a fix name for use-attribute-sets?

No, it isn't. Just as in C or Java, certain things have to be known at compile time - names of variables, functions, templates, and attribute sets are some examples. You can only use the curly-bracket notation in attributes where XSLT explicitly permits it.

Michael Kay
http://www.saxonica.com/
Author, XSLT Programmer's Reference and XPath 2.0 Programmer's Reference

micky3248 · August 18th, 2006, 03:05 AM

Another problem: I have a list in this form:

Code:

<ZMSTextarea id="e74">
<format><![CDATA[unordered_list]]></format>
<attr_dc_coverage><![CDATA[global.ger]]></attr_dc_coverage>
<text>
<lang id="ger"><![CDATA[Was ist ein Content Management System
Was ist das ZMS, was ist Zope
Wie komme ich zu so einem System fÃÂ¼r meine Abteilung
Wie lautet die Adresse
Welche Adressen kann ich verwenden 
Woher bekomme ich die Adresse
Was ist die ÃÂ¶ffentliche Adresse, was ist die Adresse fÃÂ¼r die Wartung
Wie lauten meine Login-Daten
Wie komme ich zu einem Design
Kann ich ein Design anpassen
...]]></lang></text>
<active>
<lang id="ger">1</lang></active></ZMSTextarea>

The listitems are only in a new line. How kann I tell xsl, that each line is a ney item?

Thx in advance nico

mhkay · August 18th, 2006, 03:48 AM

I can't see what the list items in your source are.

In XSLT 2.0 you can split text into lines using the tokenize() function. To see how it's done in 1.0, take a look at the str:tokenize recursive template at www.exslt.org

Michael Kay
http://www.saxonica.com/
Author, XSLT Programmer's Reference and XPath 2.0 Programmer's Reference

micky3248 · August 18th, 2006, 03:52 AM

Hi Michael,

If you have any tipps where I could look at, so I do not need to ask you each time, it would be good.

Thx for all the help. Regards Nico