You are using quotation which is assigning the variable with the string instead of the value. It should look something like this
If you place quotes around objectHeight, it will treat it as string instead of a number. That is one reason I declare the variable and define its type, this way I will get a mismatch error if I try to shove a string in a variable defined for a number.PHP Code:var objectHeight:Number = 1;
var anotherVar:Number;
anotherVar = objectHeight + 1;
trace (anotherVar) // 2;




Reply With Quote