On Sun, Feb 8, 2009 at 11:58 AM, jacmgr <[email protected]> wrote:
>
> I have 2 ideas
>
> [if more {p0} 1]
> [if* more {p0} '4']info.eznote.{p1}{p4}[if*]
> [if* equal {p0} '4']info.eznote.{p1}{p3}[if*]
> [if* equal {p0} '3']info.eznote.{p1}{p2}[if*]
> [if* equal {p0} '2']info.eznote.{p1}{p1}[if*]
> [else]
> info.eznote.{p1}{p1}
> [if]
Johns code looks more like what you are trying to do... Though we may
not be discerning exactly what you want. It seems easier to me to read
it this way though:
[if more {p0} 4]info.eznote.{p1}{p4}[if]
[if equal {p0} 4]info.eznote.{p1}{p3}[if]
[if equal {p0} 3]info.eznote.{p1}{p2}[if]
[if less {p0} 3]info.eznote.{p1}{p1}[if]
> The idea below doesn't work, but maybe someone else has clue. Anyway
> the below is just future ideas.
> Instead of all ifs, maybe one statement like:
> info.eznote.{p1}{p{p0}} //this should work but doesn't???
This doesn't write, but as my other post suggests you can do
info.eznote.{p1}{{p}:{p0}}
I'll likely change the markup rules to allow your simpler example
however. It just loosens things up.
Of course, this doesn't match the criteria above as it should be p(p0
- 1), not p(p0).
> or
> info.eznote.{p1}{p[(math {p0} - 1)]} //just future idea
This one won't work because variables are processed before functions,
so you'll end up with
info.eznote.p1{p3} (ie the 3 is inserted to late for the variables
markup to catch the {p3}.
But after a bit of testing, I think you could probably do this:
info.eznote.{p1}=[(source {p}:p{ {p0}-1 })]
Funny how some things can be a bit tricky sometimes... Love a good puzzle
Cheers,
Dan
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