Re: [xsl] Linking to a 6 level tag content

[Kahlil Johnson <jzarecta@xxxxxxxxx>]

Hi Aron,

This is the namespaces for the office:document-content

<office:document-content
xmlns:office="urn:oasis:names:tc:opendocument:xmlns:office:1.0"
xmlns:style="urn:oasis:names:tc:opendocument:xmlns:style:1.0"
xmlns:text="urn:oasis:names:tc:opendocument:xmlns:text:1.0"
xmlns:table="urn:oasis:names:tc:opendocument:xmlns:table:1.0"
xmlns:draw="urn:oasis:names:tc:opendocument:xmlns:drawing:1.0"
xmlns:fo="urn:oasis:names:tc:opendocument:xmlns:xsl-fo-compatible:1.0"
xmlns:xlink="http://www.w3.org/1999/xlink";
xmlns:dc="http://purl.org/dc/elements/1.1/";
xmlns:meta="urn:oasis:names:tc:opendocument:xmlns:meta:1.0"
xmlns:number="urn:oasis:names:tc:opendocument:xmlns:datastyle:1.0"
xmlns:svg="urn:oasis:names:tc:opendocument:xmlns:svg-compatible:1.0"
xmlns:chart="urn:oasis:names:tc:opendocument:xmlns:chart:1.0"
xmlns:dr3d="urn:oasis:names:tc:opendocument:xmlns:dr3d:1.0"
xmlns:math="http://www.w3.org/1998/Math/MathML";
xmlns:form="urn:oasis:names:tc:opendocument:xmlns:form:1.0"
xmlns:script="urn:oasis:names:tc:opendocument:xmlns:script:1.0"
xmlns:ooo="http://openoffice.org/2004/office";
xmlns:ooow="http://openoffice.org/2004/writer";
xmlns:oooc="http://openoffice.org/2004/calc";
xmlns:dom="http://www.w3.org/2001/xml-events";
xmlns:xforms="http://www.w3.org/2002/xforms";
xmlns:xsd="http://www.w3.org/2001/XMLSchema";
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance";
office:version="1.0">
<office:body>
	<office:spreadsheet>
	<table:table table:name="Sheet1" table:style-name="ta1" table:print="false">
            <table:table-column table:style-name="co1"
table:number-columns-repeated="7"
table:default-cell-style-name="Default"/>
	<table:table-row table:style-name="ro1">
	<table:table-cell office:value-type="float" office:value="23">
<text:p>23</text:p>

I actually generate an output by using this simple xsl but it seems
too raw and I have no idea what happened (which means I didnt learn at
all :p)

By simply using the xsl:
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
	<xsl:template match="office">
		<xsl:value-of select="text"/>
		</xsl:template>
</xsl:stylesheet>

But I have many doubts.
- Why I can't use the complete tag text:p or table:table-cell or office:xxx?
- Is there a way to identify it?
- How can I link all the way directly to the table:table-cell comming
from office:document (XPath)?

And yes, I am a newbie :p


On 6/23/05, Aron Bock <aronbock@xxxxxxxxxxx> wrote:
> Khalid, you don't need to put in the whole tree, but namespace
declarations,
> and syntactically complete XML helps.  As it is I've munged up your input:
>
>
> <office:document-content office:version="1.0"
>     xmlns:office="foo-office"
>     xmlns:table="foo-table"
>     xmlns:text="foo-text"
>     >
>         <office:body>
>         <office:spreadsheet>
>               <table:table >
>                    <table:table-column />
>                  <table:table-row table:style-name="ro1">
>                         <table:table-cell office:value-type="float"
> office:value="23">
>                                   <text:p>23</text:p>
>                     </table:table-cell>
>             </table:table-row>
>           </table:table>
>     </office:spreadsheet>
>     </office:body>
> </office:document-content>
>
> And this XSL:
>
> <xsl:stylesheet version="1.0"
> xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
>     xmlns:office="foo-office"
>     xmlns:table="foo-table"
>     xmlns:text="foo-text"
>     >
>
>     <xsl:output method="xml" indent="yes"/>
>
>     <xsl:template match="/">
>         <xsl:value-of select="office:document-content/office:body
>
/office:spreadsheet/table:table/table:table-row/table:table-cell
>             /text:p"
>             />
>     </xsl:template>
> </xsl:stylesheet>
>
> Outputs:
>
> <?xml version="1.0" encoding="UTF-8"?>
> 23
>
> Regards,
>
> --A
>
> >From: Kahlil Johnson <jzarecta@xxxxxxxxx>
> >Reply-To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> >To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> >Subject: Re: [xsl] Linking to a 6 level tag content
> >Date: Wed, 22 Jun 2005 23:54:35 +0300
> >
> >Sorry I meant 23.. I didn't want to put the whole tree just the
> >relevant tags what I want is to know how exactly to get to the 23
> >value.
> >
> >On 6/22/05, Aron Bock <aronbock@xxxxxxxxxxx> wrote:
> > > Khalil,
> > >
> > > I don't see a "test:p > 21" in your input.  The closest is
> > > <text:p>23</text:p>
> > >
> > > If you could post a more complete snippet, with namespace decrarations
> >and
> > > end-tags, and specify if you want top copy the element or get its text
> > > value, I assume people could respond better.
> > >
> > > Regards,
> > >
> > > --A
> > >
> > > >From: Kahlil Johnson <jzarecta@xxxxxxxxx>
> > > >Reply-To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> > > >To: XSL <xsl-list@xxxxxxxxxxxxxxxxxxxxxx>
> > > >Subject: [xsl] Linking to a 6 level tag content
> > > >Date: Wed, 22 Jun 2005 13:03:32 +0300
> > > >
> > > >I have this xml and I want to capture the content  test:p > 21 from
> > > >the XSLT however I have problems targeting the branch:
> > > >
> > > >XML: output
> > > ><office:document-content office:version="1.0">
> > > >       <office:body>
> > > >       <office:spreadsheet>
> > > >             <table:table >
> > > >                    <table:table-column />
> > > >                <table:table-row table:style-name="ro1">
> > > >                       <table:table-cell office:value-type="float"
> > > >office:value="23">
> > > >                                   <text:p>23</text:p>
> > > >
> > > >I am using xsltproc to generate an output.
>
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>


--
Kahlil Johnson
"Ya tengo GMAIL!!"