On 9/8/06, Bob Portnell <simply(_dot_)bobp(_at_)gmail(_dot_)com> 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[(_at_)item=@search]/@prog>
<xsl:variable name="prog" select="@prog" />
<xsl:for-each $scheduleSet[(_at_)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[(_at_)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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