枚举Bundle中的sks文件

Question

我想在捆绑包中列出 .sks。它们是我的游戏水平。我只想通过丢弃其余的 .sks 来了解捆绑包中有多少级别 我试过这个

let listSKS = Bundle.main.path(forResource:".//Stage1", ofType: "sks")
print(listSKS)

返回 -> ("/private/var/containers/Bundle/Application/2B72AD87-E6FE-45FF-9DCC-AB0F27F2F353/Madness Jumping.app/./Stage1.sks")

let listLevels = GameScene(fileNamed: ".//Stage1")
print(listLevels)

返回 -> 可选（名称：'Stage1' 框架：{{-896, -414}, {1792, 828}} 锚点：{0.5, 0.5}）

let listSKS2 = Bundle.main.path(forResource:".//Stage*", ofType: "sks")
print(listSKS2)

返回 -> 无

let listLevels2 = GameScene(fileNamed: ".//Stage*")
print(listLevels2)

返回 -> 无

Answer 1

您可以使用“Stage”过滤sks文件，如下所示：

let filePrefix = "Stage"
if let fileURLs = Bundle.main.urls(forResourcesWithExtension: "sks", subdirectory: nil) {
     let filteredArray = fileURLs.filter{ $0.lastPathComponent.range(of: "^\(filePrefix)", options: [.regularExpression, .caseInsensitive, .diacriticInsensitive]) != nil }
     print("Filtered Array",  filteredArray)
}

Answer 2

你应该改用urls(forResourcesWithExtension:)

let listSKS = Bundle.main.urls(forResourcesWithExtension: ".sks", subdirectory: nil)

只获取文件名：

if let listSKS = Bundle.main.urls(forResourcesWithExtension: ".sks", subdirectory: nil) {
    let files = listSKS.map(\.lastPathComponent).filter({$0.prefix(5) == "Stage"})
    print(files)
}