Re: [xsl] (simple?) xpath question

Subject: Re: [xsl] (simple?) xpath question
From: mozer <xmlizer@xxxxxxxxx>
Date: Fri, 29 Aug 2008 22:00:43 +0200
Mark,

//*[name() = 'a' or name() = 'c']

is the right syntax

Xmlizer

On Fri, Aug 29, 2008 at 9:57 PM, mark bordelon <markcbordelon@xxxxxxxxx> wrote:
> Thanks, Colin,
>
>
> Although the requirements cannot assume how many levels there are between the root node and the desired node, your solution points me to something like this. Does this work?
>
>
> //[name() == "a" and name() == "c"]
>
>
>
>
> --- On Fri, 8/29/08, Colin Paul Adams <colin@xxxxxxxxxxxxxxxxxx> wrote:
>
>> From: Colin Paul Adams <colin@xxxxxxxxxxxxxxxxxx>
>> Subject: Re: [xsl] (simple?) xpath question
>> To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
>> Date: Friday, August 29, 2008, 12:22 PM
>> >>>>> "Mark" == mark bordelon
>> <markcbordelon@xxxxxxxxx> writes:
>>
>>     Mark> All *help*!    What is the best way to query
>> xml with xpath
>>     Mark> to get a disjoint nodelist? Specifically i
>> want to include
>>     Mark> just the root node alongwith a descendent
>> node.    XML:
>>     Mark> <a>   <b>     <c>
>> </c>  </b> </a>   XPATH:   //c
>>     Mark> DESIRED RESULT NODELIST: i.e. not this:
>> <c> </c>   but
>>     Mark> rather this: <a>   <c>
>> </c> </a>
>>
>> One possibility is:
>>
>> //*[name() != "b"]
>>
>> It depends on your exact requirements.
>> --
>> Colin Adams
>> Preston Lancashire

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