RE: How to use xslt to transform xml to xml with added nodes

Subject: RE: How to use xslt to transform xml to xml with added nodes
From: Michael Teigman <michael.teigman@xxxxxxxxxxx>
Date: Wed, 22 Sep 1999 12:34:27 -0400
Thanks Mike BUT......

This is fine when the StringMap is part of the input XML. I cannot modify the
input XML. I must only transform it to produce the output XML.

Is there some way of including the StringMap in the XSL or can I create an XML
that includes the StringMap which I could import into the XSL along with  XML
that I am transforming?

     mike



                                                                  
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To:   "'xsl-list@xxxxxxxxxxxxxxxx'" <xsl-list@xxxxxxxxxxxxxxxx>
cc:    (bcc: Michael Teigman/HPG/US/Reuters)
Subject:  RE: How to use xslt to transform xml to xml with added nodes






> if textg = "xxxx", or "yyyyyyyy" or "zz" then texti = "something"
> if textg = "aaaa", or "bb" or "cccccccccccc", or "d" or "eee"
> then texti = "something else"
> [...]
> Can this be done with a transform without a huge <xsl:choose>.

You could put the string mappings in the XML, as a kind of lookup table.

<StringMap>
    <texti out="something">
        <textg>xxxxx</textg>
        <textg>yyyyy</textg>
        <textg>zz</textg>
    </texti>
    <texti out="something else">
        <textg>aaaa</textg>
        <textg>bb</textg>
        <textg>cccccccccccc</textg>
    </texti>
</StringMap>

Then you could do this, given string $textg:
<xsl:variable name="texti"
select="string(/StringMap/texti[textg=$textg]/@out)"/>


 XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list


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