Re: [xsl] Stripping payloads

[Mohit Anchlia <mohitanchlia@xxxxxxxxx>]

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/