[RESOLVED] my variable isnt registering as a variable

my variable is "objectHeight"+i
depending on what i is the variable becomes
objectHeight1
objectHeight2
objectHeight3
etc
the problem is that its not registering as a variable
edit:
by variable becomes i don't mean what its equal to but the name of the variable.

Ignore the top and the title. I'm gonna try to explain it the best i can now.
Lets say im trying to make a "targeting" system for a rpg game or something.

PHP Code:

---

if (target == 1){
    enemy_hp1 -= 20
}
if (target == 2){
    enemy_hp2 -= 20
} 


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instead of doing that for every enemy i want to do something like this. (obvoiusly it's not correct)

PHP Code:

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["enemy_hp"+target] -=20 


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How you you know its not registering?
Have you define it properly? eg: public var objectHeight:string
Have you done a trace before you add the +i to see if the problem isnt with defining the varible but with code that uses the varible?

Quote:

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Originally Posted by amcotterill

How you you know its not registering?
Have you define it properly? eg: public var objectHeight:string
Have you done a trace before you add the +i to see if the problem isnt with defining the varible but with code that uses the varible?

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i set objectHeight1 to a number
set i to 1
set another variable to be equal to "objectHeight"+i
it came out as objectHeight1 instead of the number i picked earlier.

bump

You are using quotation which is assigning the variable with the string instead of the value. It should look something like this

PHP Code:

---

var objectHeight:Number = 1;
var anotherVar:Number;

anotherVar = objectHeight + 1;
trace (anotherVar) // 2; 


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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.

Quote:

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Originally Posted by samac1068

You are using quotation which is assigning the variable with the string instead of the value. It should look something like this

PHP Code:

---

var objectHeight:Number = 1;
var anotherVar:Number;

anotherVar = objectHeight + 1;
trace (anotherVar) // 2; 


---

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.

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im not trying to add 2 vars together. i dont want objectHeight +1 i want it to be objectHeight1. i want the number not to be added to the variable but added to its name so it becomes a different variable.

bump

still no response

Okay, you can use the eval or [] commands. Depending on the version of actionscript, the [] should work in both 2.0 and 3.0. Here is an example

eval("example1") = "Hello"

or

["Example1"] = "Hello"

As long as the variables have been declared you can reference them using either of these methods. If you are using actionscript 2.0, then it won't be as much of a issue to declare the variables prior.

Did you try my suggestion above? That should do what you want?

Quote:

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Originally Posted by samac1068

Did you try my suggestion above? That should do what you want?

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It's coming up with errors

Assume you having 20 target MovieClips(target1, target2 ,...,target20) and you have to set position of all in a fetch. Using the "this" operator you can do this

for(var i=0;i<=20;i++){
this['target'+i].y=i*20;
}

as well as you can set the variable like the same

for(var i=0;i<=20;i++){
this['variable'+i] = 20;
}

wow... I fixed it

PHP Code:

---

example2 = 50;
i = 2;
this["example"+i] -= 20; 


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i had to put "this"...... I was trying to find out what was wrong for so long....

Quote:

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Originally Posted by Johnny2528

Assume you having 20 target MovieClips(target1, target2 ,...,target20) and you have to set position of all in a fetch. Using the "this" operator you can do this

for(var i=0;i<=20;i++){
this['target'+i].y=i*20;
}

as well as you can set the variable like the same

for(var i=0;i<=20;i++){
this['variable'+i] = 20;
}

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i didnt see your post when i posted earlier. thanks anyway :D