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RE: Replace a sequence of elements by an other
- To: "'xsl-list at lists dot mulberrytech dot com'" <xsl-list at lists dot mulberrytech dot com>
- Subject: RE: [xsl] Replace a sequence of elements by an other
- From: Bedwell Tom <Tom dot Bedwell at icl dot com>
- Date: Tue, 10 Jul 2001 14:28:20 +0100
- Reply-To: xsl-list at lists dot mulberrytech dot com
This would do the trick.
<xsl:template match="br">
<xsl:variable name="pos" select="position()"/>
<xsl:choose>
<xsl:when test="preceding-sibling::node()[position()=1 and
name()='br'] and preceding-sibling::node()[position()=2 and name()='br']">
</xsl:when>
<xsl:otherwise>
<br/>
</xsl:otherwise>
</xsl:choose>
</xsl:template>
-----Original Message-----
I would like to replace a sequence of elements (in input) by an other (in
output)
As shown in the following example, when i have :
- 1 br (with or without not empty preceding text node), I keep 1 br
- 2 br, I keep 2 br
- more than 2, i keep no br
<body>
my text 1 <br />
my text 2 <br /><br />
my text 3 <br /><br /><br />
my text 4 <br /><br /><br /><br />
<table>
<tr><td> <br /><br /><br /><br /><br />
other text
....
i would like to have in OUTPUT
<body>
my text 1 <br />
my text 2 <br /><br />
my text 3 <br /><br />
my text 4 <br /><br />
<table>
<tr><td> <br /><br />
other text
...
How to do this?
Cheers
Pascal Troadec
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