swift - 初始化 Struct 实例后使用 String 变量获取 Struct 属性

Question

我希望下面的代码和注释能够说明我正在尝试学习如何做。

访问名为 iphone 的结构属性，我可以，它的有效范围为:

IconSizes().iphone

访问名为 iphone 的结构属性，我不能但希望使用包含字符串值“iPhone”的变量来访问它:

IconSizes().selectedIconType

更多上下文:

selectedIconType = "iphone" // already set as String

let sizesNamesArray = IconSizes().selectedIconType // obviously raises error.

结构:

struct IconSizes {
    var typesList: Array<String>

    var iphone: [Dictionary<String, Any>]

    init() {
        self.typesList = ["iPhone"]

        self.iphone = [
            ["size":16,"name":"icon_small.png"],
            ["size":32,"name":"icon_small@2x.png"],
            ["size":32,"name":"icon_medium.png"],
            ["size":64,"name":"icon_medium@2x.png"],
            ["size":64,"name":"icon_large.png"],
            ["size":128,"name":"icon_large@2x.png"],
            ["size":128,"name":"icon.png"],
            ["size":256,"name":"icon@2x.png"]
        ]
    }
}

Answer 1

在解释语言中，在运行时引用变量名称就像它们是字符串一样通常很容易，但当您转向编译的静态语言时，就不那么容易了。明确表达可能会效果更好。

struct IconSizes {
    var typesList: Array<String>

    var iphone: [Dictionary<String, Any>]
    var types = [String : [Dictionary<String, Any>]]()

    init() {
        self.typesList = ["iPhone"]

        self.iphone = [
            ["size":16,"name":"icon_small.png"],
            ["size":32,"name":"icon_small@2x.png"],
            ["size":32,"name":"icon_medium.png"],
            ["size":64,"name":"icon_medium@2x.png"],
            ["size":64,"name":"icon_large.png"],
            ["size":128,"name":"icon_large@2x.png"],
            ["size":128,"name":"icon.png"],
            ["size":256,"name":"icon@2x.png"]
        ]
        self.types["iPhone"] = self.iphone
    }
}

let selectedIconType = "iPhone"
print(IconSizes().types[selectedIconType])