CODEWITHSUNDEEP
arraysactionscript-3loopschildren
Anonymous
Anonymous

Reputation: 1893

Get an array of specific added children

I'm using the following array to add children to the stage:

for(var i=0;i<6;i++) {
    var aCherry=new cCherry()
    aCherry.y=10
    aCherry.x=10+100*i
    stage.addChild(aCherry)
}

Now I want to modify each cherry based on another array. Something like this:

var cherryLetter:Array=[1,0,0,0,0,0]
    for(i=0;i<6;++) {
        if(cherryLetter[i]) stage.getChildByName("aCherry")[i].y+=90
    }

Clearly stage.getChildByName("aCherry")[i] isn't correct, but coming from JavaScript this makes the most sense to me and should accurately portray what I'm trying to achieve for you guys reading this. So, how would I actually do this? This being getting an array of children added to the stage under a certain name or class (so an array of cCherry would work too, if necessary), then using them in a way similar to the above loop.

Upvotes: 0

Views: 1045

Answers (1)

Adam Smith
Adam Smith

Reputation: 1937

Here is my recommendation for how the code might look, based on the desire to use getChildByName() to find the instances of your cCherry class. Please note that I've changed the class name to Cherry in the example (which I recommend, since capitalizing class names is AS3 convention). Also, it's good practice to end statements with semi-colons. While it's usually optional, there are cases where omitting the semi-colon can produce very difficult to track down runtime bugs, so I recommend getting int he habit of using them. I also recommend including type in all your variable declarations, as shown with var aCherry:Cherry, for example.

var i:int;
for(i=0; i<6; ++i)
{
    var aCherry:Cherry=new Cherry(); // Note, it's my recommendation that you rename cCherry class to Cherry (convention)
    aCherry.y=10;
    aCherry.x=10+100*i;
    aCherry.name = "aCherry" + String(i); // String() cast for clarity only, not necessary
    stage.addChild(aCherry);
}

and

var cherryLetter:Array=[1,0,0,0,0,0];
for(i=0; i<6; ++i)
{
    var cherry:Cherry = stage.getChildByName("aCherry" + String(i)) as Cherry;
    if(cherry && cherryLetter[i]) cherry.y += 90;
}

Upvotes: 1

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