Subject: Re: [xsl] Stripping payloads From: David Carlisle <davidc@xxxxxxxxx> Date: Mon, 22 Aug 2011 23:32:07 +0100 |
I am wondering if it's even possible to get 2 separate output from single parse?
I was able to get output 1 by doing something like:
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="@* | node()"> <xsl:copy> <xsl:apply-templates select="@* | node()"/> </xsl:copy> </xsl:template>
<xsl:template match="payload"> <payload></payload> </xsl:template>
</xsl:stylesheet>
On Mon, Aug 22, 2011 at 1:20 PM, Mohit Anchlia<mohitanchlia@xxxxxxxxx> wrote:Sorry I didn't specify it correctly. I need to extract just the payload value. So output would like something like:
output 1 (no payload value): <api> <name>get</name> <payload></payload> </api>
output 2 (just the payload value):
assaddddd
On Mon, Aug 22, 2011 at 1:15 PM, Mohit Anchlia<mohitanchlia@xxxxxxxxx> wrote:On Sun, Aug 21, 2011 at 10:46 AM, Martin Honnen<Martin.Honnen@xxxxxx> wrote:Mohit Anchlia wrote:
I have payload something like
<api> <name>get</name> <payload>assaddddd</payload> </api>
<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:template match="api"> <xsl:result-document href="a.xml"> <api> <xsl:copy-of select="* except payload"/> </api> <xsl:result-document> <xsl:result-document href="b.xml"> <api> <xsl:copy-of select="payload/node()"/> </api> <xsl:result-document> <xsl:template> </xsl:stylesheet>
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