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有没有办法让 Apple 的genstrings命令行工具识别从SwiftUI的LocalizedStringKey初始化程序定义的可本地化字符串键？

对于这个输入文件 ( testing-genstrings.swift): ...

import UIKit
import SwiftUI

enum L10n {
    static let test0 = NSLocalizedString("TEST0", comment: "")

    static let test1 = LocalizedStringKey("TEST1")

    static func test2(_ parameter: String) -> LocalizedStringKey {
        return LocalizedStringKey("TEST2_\(parameter)")
    }

    static func test3(_ parameter: String) -> String {
        return NSLocalizedString("TEST3_\(parameter)", comment: "")
    }

    static func test4(_ parameter: String) -> String {
        return String.localizedStringWithFormat(NSLocalizedString("TEST4", comment: ""), parameter)
    }
}

let test5 = "TEST5"
let test6 = "TEST6"
let test7 = "TEST7"

struct TestView: View {
    var body: some View {
        VStack {
            Text(L10n.test0)
            Text(L10n.test1)
            Text(L10n.test2("foo"))
            Text(L10n.test3("bar"))
            Text(test5)
            Text(LocalizedStringKey(test6))
            Text(NSLocalizedString(test7, ""))
            Text("TEST8")
            Text("TEST9_\("baz")")
        }
    }
}

...genstrings生成此输出：

$ genstrings -SwiftUI -s LocalizedStringKey testing-genstrings.swift && iconv -c -f utf-16 -t utf-8 Localizable.strings
genstrings: error: bad entry in file testing-genstrings.swift (line = 9): Argument is not a literal string.
genstrings: error: bad entry in file testing-genstrings.swift (line = 11): Argument is not a literal string.
genstrings: error: bad entry in file testing-genstrings.swift (line = 12): Argument is not a literal string.
genstrings: error: bad entry in file testing-genstrings.swift (line = 36): Argument is not a literal string.
genstrings: error: bad entry in file testing-genstrings.swift (line = 37): Argument is not a literal string.
genstrings: error: bad entry in file testing-genstrings.swift (line = 37): Argument is not a literal string.
/* No comment provided by engineer. */
"bar" = "bar";

/* No comment provided by engineer. */
"foo" = "foo";

/* No comment provided by engineer. */
"TEST0" = "TEST0";

/* No comment provided by engineer. */
"TEST3_\(parameter)" = "TEST3_\(parameter)";

/* No comment provided by engineer. */
"TEST4" = "TEST4";

/* No comment provided by engineer. */
"TEST8" = "TEST8";

/* No comment provided by engineer. */
"TEST9_%@" = "TEST9_%@";

您可以看到它识别通过NSLocalizedString和Text的初始化程序 Text() 使用ExpressibleByStringInterpolation(TEST9_%@在示例中) 的初始化程序定义的键，但在使用 定义的所有键上都失败了LocalizedStringKey。

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1 回答 1

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genstrings比较幼稚。它正在寻找一个有两个参数的函数，第一个未命名，第二个命名为“comment”。

如果您添加了以下扩展名：

public extension LocalizedStringKey {
    init(_ value: String, comment: String) {
        self.init(value)
    }
}

并且总是使用它，你可以LocalizedStringKey通过传递-s LocalizedStringKey给genstrings.

请记住，如果您声明LocalizedStringKey为返回类型或变量，也会产生genstrings错误。所以你需要一个单独typealias LocalizedStringKeyResult = LocalizedStringKey的，你在引用时使用LocalizedStringKey但不想genstrings抱怨。

而且，当然，你不会得到你想要的插值，因为genstrings它只适用于Text.

真正的答案是......不要使用LocalizedStringKey. 尽可能使用Text（以获得插值）。NSLocalizedString不能的时候用。

于 2021-02-03T01:26:32.720 回答