CODEWITHSUNDEEP
actionscript-3
Andy Moore
Andy Moore

Reputation: 865

getDefinitionByName not finding a variable that exists

I am using flash.utils.getDefinitionByName in an attempt to grab an art asset. I use this function quite a bit and haven't had trouble until now. Check it:

assetName = Assets.MegaBerry; // works
assetName = getDefinitionByName("Assets.MegaBerry") as Class; // doesn't work

What the heck?? Error response for the second line is "Variable not found."

If it matters: Assets is a file in my root source directory (it has no package; Assets is the fully qualified name) and I've tried putting:

import Assets;

at the top with no luck.

For reference, in Assets.as I have:

    [Embed(source = "../art/Inventory/MegaBerry.png")]
    public static var MegaBerry:Class;

Upvotes: 3

Views: 1038

Answers (5)

Michael Antipin
Michael Antipin

Reputation: 3532

If you want to find out what the fully qualified name of you class really is, you may do the following:

trace(getQualifiedClassName(Assets.MegaBerry));

You may do that from inside Assets.as, for instance.

You can feed that string back to getDefinitionByName() and get a reference to the class.

trace(getDefinitionByName(getQualifiedClassName(SomeClass)));
// output [class SomeClass]

And remember, getDefinitionByName() only gets you references for classes that are in the same scope as the getDefinitionByName call itself. So, if you are loading external SWFs, getting class references will depend on the application domain you are using and the place, where this code executes.

Upvotes: 0

weltraumpirat
weltraumpirat

Reputation: 22604

Your problem is that embedding the resource into the Assets class will create a static variable of type Class that belongs to that class - which is what you are referencing when you use Assets.MegaBerry: A variable(!) of type Class.

It does not, however, register the MegaBerry class to a fully qualified class name. To do this, you have to use - who would have guessed it - registerClassAlias at some point in your application:

registerClassAlias("Assets.MegaBerry", Assets.MegaBerry);

After that, it will be available everywhere else when calling getDefinitionByName.

** EDIT **

Well that's some unexpected behavior... It turns out, the class that was embedded is in fact automatically registered, but under {className}_{variableName}, instead of the notation you would expect. So using:

getDefinitionByName("Assets_MegaBerry") as Class;

should to the trick.

registerClassAlias also works, but then you need to call getClassByAliasinstead of getDefinitionByName. Sorry for the mix-up.

** END EDIT **

You can also use the Embed tag to inject the resource into a separate class file, which you can then reference as expected by using getDefinitionByName, or simply using an import:

package assets {

    [Embed(source="../art/Inventory/MegaBerry.png"]
    public class MegaBerry extends BitmapData {
    }
}

Upvotes: 3

xeorem
xeorem

Reputation: 593

In actionscript, objects actually have a name property that is different from the actual variable name as it shows in code.

For example, if you create a variable as follows, 

var myBerry = new MegaBerry();

Then getDefinitionByName("myBerry") will return null.

Only when you set the name of the variable by writing myBerry.name = "myBerry", will getDefinitionByName("myBerry") return what you want it to. The name of the object doesn't necessarily have to be equal to the variable name in code.

In your specific case, I don't think you need to use any of that anyways. Have you tried assetName = new MegaBerry() ?

Upvotes: 0

Andy Moore
Andy Moore

Reputation: 865

Instead of calling

assetName = getDefinitionByName("Assets.MegaBerry") as Class;

, instead just use:

assetName = Assets["MegaBerry"];

Upvotes: 1

GGG
GGG

Reputation: 817

try:

 [Embed(source = "../art/Inventory/MegaBerry.png" , symbol="MegaBerry")]
 public static var MegaBerry:Class;

Upvotes: 0

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