Subject: Re: [xsl] Stripping payloads From: Mohit Anchlia <mohitanchlia@xxxxxxxxx> Date: Mon, 22 Aug 2011 14:57:50 -0700 |
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> >>>> >>>> We have got a requiement to save all incoming request except the >>>> "payload" on the wire. Also, one of the option I think is to use XSL >>>> transformation to strip the payloads. I need some suggestion on if >>>> it's the good way and also how can I possibly do it using xslt. >>> >>> Well if you want to strip the payload element(s) then do >>> >>> <xsl:template match="@* | node()"> >>> <xsl:copy> >>> <xsl:apply-templates select="@* | node()"/> >>> </xsl:copy> >>> </xsl:template> >>> >>> <xsl:template match="payload"/> >> >> In the same transformation is it possible to output 2 separate pieces: >> >> 1) Payload >> 2) Everything but payload >> >> We need to persist them but separately >> >>> >>> >>> >>> >>> -- >>> >>> Martin Honnen --- MVP Data Platform Development >>> http://msmvps.com/blogs/martin_honnen/
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