Re: [xsl] Process the following group

["Wendell Piez wapiez@xxxxxxxxxxxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx>]

Hi again,

With the bug repaired --

<xsl:key name="myDivs" match="div2" use="(preceding-sibling::div1[1],
following-sibling::div1)[1]/generate-id()"/>

Lesson? XPath gives tremendous power enabling workarounds for even pesky
exception conditions.

Cheers, Wendell


On Fri, Jul 8, 2022 at 10:09 AM Wendell Piez wapiez@xxxxxxxxxxxxxxx <
xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> wrote:

> Just writing again to note there's a bug in the code I just posted! :->
>
>
> On Fri, Jul 8, 2022 at 10:03 AM Wendell Piez wapiez@xxxxxxxxxxxxxxx <
> xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> wrote:
>
>> Rick,
>>
>> That being said, this can also be done in one pass, and indeed without
>> xsl:for-each-group at all ....
>>
>> <xsl:key name="myDivs" match="div2" use="(preceding-sibling::div1,
>> following-sibling::div1)[1]/generate-id()"/>
>>
>> <xsl:template match="div1">
>>   <xsl:copy>
>>     <xsl:apply-templates select="key('myDivs',generate-id())"
>> mode="include"/>
>>   </xsl:copy>
>> </xsl:template>
>>
>> etc.
>>
>> This works on the principle that groups can be created by relating
>> elements together using keys (old 1.0 approach).
>>
>> Not disagreeing with Eliot's advice here however -- splitting into
>> multiple passes will often yield a more maintainable as well as clearer
>> solution to a problem. A solution like this is both more obscure, and more
>> apt to break when something subtle turns up.
>>
>> Cheers, Wendell
>>
>>
>>
>> On Fri, Jul 8, 2022 at 9:30 AM rick@xxxxxxxxxxxxxx <
>> xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> wrote:
>>
>>> Hi Eliot, thank you for walking me through the logic. It makes sense so
>>> I will try to code it that way. But I think the advice from you and
others
>>> to do a multiple pass may be a more robust approach. Thanks to all that
>>> responded. -Rick
>>>
>>>
>>>
>>> *From:* Eliot Kimber eliot.kimber@xxxxxxxxxxxxxx <
>>> xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx>
>>> *Sent:* Thursday, July 7, 2022 6:11 PM
>>> *To:* xsl-list@xxxxxxxxxxxxxxxxxxxxxx
>>> *Subject:* Re: [xsl] Process the following group
>>>
>>>
>>>
>>> So the first two div2 elements should be grouped with the first div1?
>>>
>>>
>>>
>>> If you do group-starting-with div1 then the first two div2s will be in
>>> the first group and that group will not start with a div1, so you can
have
>>> logic to ignore that first group and then in a group that starts with
>>> <div1>, you can pull any <div2> elements that do not have a <div1>
>>> preceding sibling, which will get the <div2>s that would have formed the
>>> first group. No subsequent group will have <div2>s that do not have a
>>> <div1> preceding sibling.
>>>
>>>
>>>
>>> But this is also a case where it might be easier to pre-process your
>>> content before grouping, for example, to move things so that therebs
always
>>> a <div1> before any <div2>s or what ever it might be.
>>>
>>> I generally find these kinds of challenges are most effectively solved
>>> by breaking them into several steps, each one of which is relatively easy
>>> to implement.
>>>
>>>
>>>
>>> Cheers,
>>>
>>>
>>>
>>> E.
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>>>
>>
>>
>> --
>> ...Wendell Piez... ...wendell -at- nist -dot- gov...
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>
>
> --
> ...Wendell Piez... ...wendell -at- nist -dot- gov...
> ...wendellpiez.com... ...pellucidliterature.org... ...pausepress.org...
> ...github.com/wendellpiez... ...gitlab.coko.foundation/wendell...
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--
...Wendell Piez... ...wendell -at- nist -dot- gov...
...wendellpiez.com... ...pellucidliterature.org... ...pausepress.org...
...github.com/wendellpiez... ...gitlab.coko.foundation/wendell...