xJSFL

sstark

Hello,

I couldn't get the selector $$ to work with library fonts, I had to instead iterate through the entire library and look for all instances that are fonts.

I feel like I am doing this the hard way, or I am missing something?

Tries:

$$("fonts");

but instead, need to use:

var libraryItems = dom.library.items;
var i = libraryItems.length;
while (i--) {
	var item = libraryItems[i];
	if (item.itemType == "font") {
		processFont(item);
	}
}

Dave

Sorry for the delay in replying... the forum's been a bit quiet for a bit!

It's:

$$(':font');

That's a pseudo (filter) class, just like CSS. In this case, looking for an item, the type of which, is "font".

You could also pick ":movieclip", ":graphic", etc. as well as the other pseudo properties like ":last", ":random", or ":children".

Back to your syntax, as you've just done this...

$('fonts');

...xJSFL would look for any items named "fonts", so "assets/fonts" or such like, would be found.

There's a handy table here with the all the syntaxes: http://www.xjsfl.com/support/api/elements/selectors#selectors-table

sstark

I tried a few things but it didn't work, I'll try your suggestions and let you know. thanks!

Dave

Any reason it wasn't clear in the first place? Are the docs not up to scratch?

xJSFL

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