Re: [xsl] A challenge.. Group Periods of Data (1..5, 2..8, 4..9) (10..12; 10..14

["Aron Bock" <aronbock@xxxxxxxxxxx>]

As they say, those who can, do, and those who can't, teach.  Thus perhaps I 
could annotate the solution below:


> From: Karl Stubsjoen <kstubs@xxxxxxxxx>
>
> 1)  Variable e2 select.
> select="@period_end[. &gt; $e]|$e[. &gt;= current()/@period_end]"
>
> How does the pipe work here and is this only evaluating the current B
> element, or evaluating all @period_end(s)?

$e is a passed-in param; so this expression selects either current 
<B>/@period_end if greater than $e, or simply the passed-in $e if it's 
equal-or-greater than the current <B>/@period_end.  Remember that the pipe, 
"|", indicates "union", not "or" as in many programming languages.


> 2)  Right off the bat (first iteration), I don't understand how you
> determine the period attribute "ends" value.

It assumes the period ends with the @period_end of the first <B>, until 
proven wrong further downstream.

> 3)  Variable g select, what does this get you, the ancestor record?
> select="/.."