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Re: [xsl] Joining sibling elements
Subject: Re: [xsl] Joining sibling elements From: Mukul Gandhi <gandhi.mukul@xxxxxxxxx> Date: Thu, 11 Aug 2005 12:07:46 +0530 |
Given this XML - <?xml version="1.0" encoding="UTF-8"?> <root> <a><b>this</b><b>this 1</b><b>this 2</b></a> <a><b>this 3</b></a> <a><b>this 4</b><b>this 5</b></a> <a><b>this 6</b><b>this 7</b><b>this 8</b><b>this 9</b></a> <a><b>this 10</b><b>this 11</b><b>this 12</b><b>this 13</b></a> </root> This XSLT stylesheet - <?xml version="1.0" encoding="UTF-8"?> <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0" xmlns:common="http://exslt.org/common" exclude-result-prefixes="common"> <xsl:output method="xml" indent="yes"/> <xsl:template match="/root"> <root> <xsl:apply-templates select="a"/> </root> </xsl:template> <xsl:template match="a"> <b> <xsl:variable name="rtf"><xsl:for-each select="b"><xsl:value-of select="."/><xsl:text> </xsl:text></xsl:for-each></xsl:variable> <xsl:value-of select="common:node-set($rtf)/text()"/> </b> </xsl:template> </xsl:stylesheet> Produces following output - <?xml version="1.0" encoding="utf-8"?> <root> <b>this this 1 this 2 </b> <b>this 3 </b> <b>this 4 this 5 </b> <b>this 6 this 7 this 8 this 9 </b> <b>this 10 this 11 this 12 this 13 </b> </root> Hope this helps, Regards, Mukul On 8/10/05, Marcin Mi3kowski <milek_pl@xxxxx> wrote: > Hi, > > I'm wondering if it's practical to use XSLT in a following scenario: you > have an XML file which should be reformatted (formatted "tidy") to join > sibling elements, like: > > <b>this</b><b> </b><b>is</b><b> </b><b>bold<text</b> > > which should be transformed this: > > <b>this is bold text</b> > > Or, in a more complex example: > > <a><b>this</b></a><a><b> </b></a><a><b>more</b></a><a><b> > </b></a><a><b>text</b></a> > > So this is a transformation to keep the resulting file tidy, with less > tagging. I'm planning such a transformation for WordML files before > submitting them for translation where less tagging means higher > translation quality. > > Thank you for all advice, > Marcin Milkowski
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