Expression concatenation + random

titos

I have attribute named game.RandomPick of type text that contain a random name of a attribute.
It contain text that is random: a1, a2 or a3

I want a display the content of those attribute a1, a2 or a3. So I try this expression in vain.

`game..(game.RandomPick)`

gyroscope

Hi titos, presumably you mean in a Display Text Behaviour? Just put game.RandomPick. :-)

titos

game.randomPick will give me ex: a1
But A1 is attribute that contain text, and I want to show the content of the attribute A1.

JeffreyShimane

You could do something like this...

Create a game attribute called RandomNumber (integer type).

Change Attribute game.RandomNumber to random(1,3)

If game.RandomNumber = 1,
Display Text game.A1 (or you could just have the contents of A1 written here)

If game.RandomNumber = 2,
Display Text game.A2(or you could just have the contents of A2 written here)

If game.RandomNumber = 3,
Display Text game.A3 (or you could just have the contents of A3 written here)

- Jeff

titos

Thanks, I am sure there is another way.

CodeMonkey

"game"..game.RandomPick

Text needs to be in quotation marks. Expressions and attributes do not, but must be separated by '..' between each concatenated section.

titos

I am trying to get a random number between 1-3.
Then I have 3 attribute called a1,a2 and a3 that contain text:a1 => "allo" , a2=> "mon", a3 => "coco".
So when I get my random 1 2 or 3 it represent the numbers in my attribute name. letter "a" return in every attribute name. That number is stored in the attribute game.randomPicked.

So I want when the random number is 1 to concatenate a with 1 to make a1. "a"..game.randomPicked

Then I want to ask how to get the content of game.a1 without a if.