Question

我正在尝试创建一个String扩展来执行类似的操作

"My name is %@. I am %d years old".localizeWithFormat("John", 30)

看起来像这样

extension String {
  func localizeWithFormat(arguments: CVarArgType...) -> String {
    return String.localizedStringWithFormat(
      NSLocalizedString(self,
        comment: ""), getVaList(arguments))
  }
}

它给我以下编译错误

类型CVaListPointer不符合协议CVargType

任何人都知道如何解决这个编译错误？

Answer 1

这应该非常简单，只需按照以下步骤更改参数：

extension String {
    func localizeWithFormat(name:String,age:Int, comment:String = "") -> String {
        return String.localizedStringWithFormat( NSLocalizedString(self, comment: comment), name, age)
    }
}

"My name is %@. I am %d years old".localizeWithFormat("John", age: 30)  // "My name is John. I am 30 years old"

init(format:locale:arguments:)

extension String {
    func localizeWithFormat(args: CVarArgType...) -> String {
        return String(format: self, locale: nil, arguments: args)
    }
    func localizeWithFormat(local:NSLocale?, args: CVarArgType...) -> String {
        return String(format: self, locale: local, arguments: args)
    }
}
let myTest1 = "My name is %@. I am %d years old".localizeWithFormat(NSLocale.currentLocale(), args: "John",30)
let myTest2 = "My name is %@. I am %d years old".localizeWithFormat("John",30)

Answer 2

这允许带有可变参数的本地化字符串：

extension String {
      func localizedStringWithVariables(vars: CVarArgType...) -> String {
        return String(format: NSLocalizedString(self, tableName: nil, bundle: NSBundle.mainBundle(), value: "", comment: ""), arguments: vars)
      }
}

使用以下方式致电：

"Hello, %@. Your surname is: %@.".localizedStringWithVariables("Neil", "Peart")

Answer 3

在Swift 3中

func localize(key: String, arguments: CVarArg...) -> String {
  return String(format: NSLocalizedString(key, comment: ""), arguments)
}

Swift中的localizeWithFormat和variadic参数

3 个答案: