Subject: Re: [xsl] xtvd grouping problem (I think) From: "Bob Portnell" <simply.bobp@xxxxxxxxx> Date: Fri, 8 Sep 2006 14:25:55 -0700 |
Let me clarify a little. I already have my set of programs in a variable... call it $hitSet. So that's cool. And the bit from the sort on down I'd already worked out. Good and good.
I need to get away from any looping based on program IDs. That's my problem: program IDs are more fine-grained than my search item terms. So I need to expand up, collecting up all the <schedule>s which have the program IDs from $hitSet (and so representing each search item discovery individually)
In pseudocode, my vision is something like for each member of schedule ( a long list) if this member's @program is found among the @ids in (hitSet list) push (this member into Keepers list)
But I have no idea if that's possible in XSLT. I have a hunch it is, and I'm just overlooking a very simple XPath function for the assignment...
Bob P simply.bobp@xxxxxxxxx
On 9/8/06, Bob Portnell <simply.bobp@xxxxxxxxx> wrote: > for each search item > find the set of programs which have that item > convert this to the set of schedule items which have those programs > for each in the schedule set > sort by time > report the show information.
<xsl:for-each $progSet/items[@item=@search]/@prog> <xsl:variable name="prog" select="@prog" /> <xsl:for-each $scheduleSet[@prog = $prog]/@prog> <xsl:sort select="@time" /> <xsl:value-of select="@showInfo" </xsl:for-each> </xsl:for-each>
> Given a variable $progSet which has a bunch of <programs> in it, how > do I define a $schedSet which has only <schedule> items containing > @program ids from $progSet? > > I've gotten as far as > <xsl:variable name="schedSet" select="//schedule[@program . . {sound > of screeching brakes}? />
Pretty sure I touched on this above. Let me know.
> Please excuse where I've strayed from proper XSLT vocabulary; I hope > my intent is clear enough despite such lapses of inexperience.
No problemo. Was very well conveyed. =).
-S
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