Boolian's maxed out or......specific time & date

Support
1

I want to show an element on a specific month/day/time. Using && I can get month (or day), hour, minute but not all 4
for example this works:
$#DM#==7&&#DUH#==19&&#Dm#==32?100:0$
and this does not:
$#DM#==7&&#Dd#==16&&#DUH#==19&&#Dm#==32?100:0$
It doesn’t work as far as I know because of the limit of 3 Boolian operators. I also fell into the brackets trap but figured it out eventually!
Seems like a simple request to make something happen at a certain date and time but I can’t figure it out.

I have a question about using #DUH# but I’ll put that in a separate post.

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2

Ha ha yes . What fun . Now first off you could use the Day number in the year #DD# to replace the first two . You could use #DWFHS# and work out which angle your time was . I struggle with multi conditionals so I split them up . So you could use half your formula to send it off stage X and the other half off stage Y .

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I did not test it on watch, and it may not work reliably for all models, but in preview this works for me:
(100*(#DM#==7)*(#Dd#==16)*(#DH#==19)*(#Dm#==32))

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Nice one Peter . I did not realise WFS Voodoo worked on here . Why would 4 )*( work and not 4 && .
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So Peter’s Voodoo Conditional Worked fine on my GW4 Classic . Master class Bookmarked .
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5

Does #DD# take care of leap year - is it smart enough to know it’s a leap year and move the number of days to keep my date if it’s after the 29th of February?

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Wow, heros

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There is a good point . Without testing it I am not sure . Yes you are right . Sorry . There is a Tag or the number of days in the year .
Try the Peter Voodoo . You realise your formula Lights something for 1 minute .
((#DIY#)-#DD#) is the same Leap Year or no but you have to work backwards . The time machine gives you the correct number .

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Yes I know! This is only half the story and I have to try and figure the rest out myself. Can’t make things too easy for myself!

The other half is to make the element (s) stay for a period of time before disappearing…

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9

It is always good to reduce the number of conditions down to 2.

So I was thinking, why not just add up the numbers.

3rd of May 19:35 would be: #DM#+#Dd#+#DH#+#Dm#
which equal to 62.

However, 62 would also be valid on the 5th of March, so that brings us to a second condition. How about using the month?

So the opacity formula would be:
$(#Dd#+#DM#+#DH#+#Dm#)=62&&#DM#=5?100:0$

Is my thinking right? or did I miss something?

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10

It counts days of year from 1 to 365 or 366 respectively, so your date after 29th feb. would cause the result being off 1 day on leap year.

Counting hours and minutes together would make the condition met many times a day.

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11

Did you see the formula for the day in the year .

((#DIY#)-#DD#)

.Think about using the Hour angle #DWFHS# for the time conditional .

$#DWFHS#>=90&&#DWFHS#<=270?100:0$

Would show between 6am and 6 pm . As before use the time machine to get your angle .

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12

try:

$#DM#*31+#Dd#==7*31+16&&#DUH#*24+32==19*24+32?100:0$
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try this:

instead of
$#DM#==7&&#Dd#==16&&#DUH#==19&&#Dm#==32?100:0$

just concatenate the parts, but take care of leading zeroes then:

$#DM##DdL##DUHZ##DmZ#=7161932?100:0$