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I have a movieclip without name, and the library name is "foo".

But if I don't use the option "Export for ActionScript", the movieclip class is "flash.display.MovieClip", and I can't find using the library name "foo" like

loop over all of the elements: if (getQualifiedClassName(this.getChildAt(i)) == "foo")

and the .name propertie is "instance1".

There is a way to identify a movieclip "class" or library name without using the "Export for ActionScript"? Thanks.

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2 Answers

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it's very easy to add a class name (double click on the empty field under "AS Linkage" in the library and type a name). if you don't give it a class any instances will be instances of MovieClip (extends Sprite). if you do give it a class, any instances will be instances of that class (extends MovieClip). you can verify by typing something like trace(describeType(getChildAt(0)));.

the library and folder names are purely organizational. you can find them using your JSFL script because JSFL is talking to the IDE which is doing the organizing... sorry

if for some reason you can't add a class name, maybe you know something else about the instance in question? like it has a child named something specific (which you can search for).

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Thanks, I will try to add a warning about it in my application. –  Tiago Peczenyj Nov 24 '12 at 17:29
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To my knowledge if you don't export a class for ActionScript it will be initialized as a generic DisplayObject with no properties set other than the graphics. You won't be able to check for it using name or the type of the object because it has no name or type to check against.

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ok, but I can find the library name using a jfx script –  Tiago Peczenyj Nov 22 '12 at 16:35
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