"OR" Using the expression editor?
3absh
Member Posts: 601
I want to write this:
When attribute X = 1, or 2
but I need to write it using the expression editor not the "When ANY conditions are valid"
which of the expression editor function does this?
Comments
You could use any.
When Any
If x = 1
If x = 2
Or like this
When All
If x >= 1 and x <= 2
Why does it need to be written in the expression editor/text box?
(Game.Attribute==1)and(DOACTIONFOR1)or(Game.Attribute==2)and(DOACTIONFOR2)or(DOACTIONIFNONEAREMET)
I use this every now and then. Not sure if its most practical solution though.
If numeric expression: x-round(x/4) = 1
. . . then x must be 1 or 2.
Another way:
If numeric expression: ceil(x/2) = 1
. . . then x must be 1 or 2.
Using Numeric Expression:
(game.X==1)or(game.X==2) = 1
The = 1 is the built in = and the 1 is in the right hand box. It does exactly the same as @KevinCross suggested, though. There's no advantage to this over his way.
thats a lot of solutions
There are probably an infinite amount of solutions to this (I think).
Here's another
If numeric expression: abs(floor(x*1.5)-2) = 1
. . . then x must be 1 or 2.
Here's another (somebody stop me !!
)
If numeric expression: abs(((x-1)*2)-1) = 1
. . . then x must be 1 or 2.
When numeric expression textFind("#1#2","#"..self.x,1) ≠ -1
@tatiang Love it! Another (not as good) textFind option:
max(textFind(self.X,"1",1),textFind(self.X,"2",2)) = 1
(This breaks pretty easily though, if the conditions are too wide. Like, say, 11. That can be fixed by throwing two min(x) in there though.)
max(textFind(min(self.X,3),"1",1),textFind(min(self.X,3),"2",2)) = 1
Very similar to one @Socks already did, but different enough!
round(sqrt(self.X)) = 1
I see your square root and raise you a sine function:
ceil(sin(min(( self.X *90)-1,181))) = 1
@Socks I don't think anyone is going to beat a sin() one. One more anyway, though! We've not had any where the result has to be 2 yet.
ceil((self.X*5)/10+1) = 2
Wow, I'm overwhelmed.
Thanks guys!!
OMG I just noticed that what I actually needed was
0 OR 1
not
1 OR 2
Ultimate FAIL on my behalf.
That makes it much easier. If X can only be positive numbers, just say x > 1.
If X can be negative, it gets a little more complicated but it still very much doable.
Then 3 (and 7,928 and 4.32 . . etc) would pass the condition ? But 0 and 1 would fail ? Maybe I misunderstand what you are saying, but either way this is highly controversial maths
if abs((X*2)-1) =1
Only 0 and 1 will pass this condition.
Or (as Armelline says) if you are only using positive integers then you can simply check for x < 2.
oop, I mean < 2. No idea why I said > 1
I must have transposed his question to NOT 0 or 1...