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Idea: load()로 분산된 파일들 얻어오기 #1

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ghost opened this issue Feb 25, 2015 · 1 comment

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ghost commented Feb 25, 2015

load("Zipper.js"); 로 로드하고(디렉토리는 알아서)
Zipper.js의 내용은 Pool.Zipper = {~~~} 이러한 형식 어떨까요

이렇게 하면 object별로 분산 관리가 되어서 편합니다

ghost added the idea/vote label

ghost referenced this issue


          Entity객체 생성 및 메서드 추가

fb0170a

Contributor

jaehee0507 commented Feb 26, 2015

load는 어떻게 쓰는 건가용?

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