斯威夫特，斯威夫特

有不同的情况下，从一个数据类型转换到另一个数据类型，这取决于输入。

如果输入数据类型是Any，我们必须像以前一样使用convert到实际的数据类型，然后转换为我们想要的数据类型。例如:

func justGetDummyString() -> Any {
  return "2000"
}
let dummyString: String = (justGetDummyString() as? String) ?? "" // output = "2000"
let dummyInt: Int = Int(dummyString) ?? 0 // output = 2000

有用的字符串到Int和其他类型

extension String {
        //Converts String to Int
        public func toInt() -> Int? {
            if let num = NumberFormatter().number(from: self) {
                return num.intValue
            } else {
                return nil
            }
        }

        //Converts String to Double
        public func toDouble() -> Double? {
            if let num = NumberFormatter().number(from: self) {
                return num.doubleValue
            } else {
                return nil
            }
        }

        /// EZSE: Converts String to Float
        public func toFloat() -> Float? {
            if let num = NumberFormatter().number(from: self) {
                return num.floatValue
            } else {
                return nil
            }
        }

        //Converts String to Bool
        public func toBool() -> Bool? {
            return (self as NSString).boolValue
        }
    }

像这样使用它:

"123".toInt() // 123

基本想法，注意这只适用于Swift 1。(查看ParaSara的回答，看看它是如何在Swift 2.x中工作的):

    // toInt returns optional that's why we used a:Int?
    let a:Int? = firstText.text.toInt() // firstText is UITextField
    let b:Int? = secondText.text.toInt() // secondText is UITextField

    // check a and b before unwrapping using !
    if a && b {
        var ans = a! + b!
        answerLabel.text = "Answer is \(ans)" // answerLabel ie UILabel
    } else {
        answerLabel.text = "Input values are not numeric"
    }

Swift 4更新

...
let a:Int? = Int(firstText.text) // firstText is UITextField
let b:Int? = Int(secondText.text) // secondText is UITextField
...

斯威夫特4.0

let stringNumber = "123"
let number = Int(stringNumber) //here number is of type "Int?"


//using Forced Unwrapping

if number != nil {         
 //string is converted to Int
}

你也可以使用可选绑定而不是强制绑定。

eg:

  if let number = Int(stringNumber) { 
   // number is of type Int 
  }

你可以使用NSNumberFormatter(). numberfromstring (yourNumberString)。这很好，因为它返回一个可选的，然后你可以用if let测试，以确定转换是否成功。 如。

var myString = "\(10)"
if let myNumber = NSNumberFormatter().numberFromString(myString) {
    var myInt = myNumber.integerValue
    // do what you need to do with myInt
} else {
    // what ever error code you need to write
}

斯威夫特5

var myString = "\(10)"
if let myNumber = NumberFormatter().number(from: myString) {
    var myInt = myNumber.intValue
    // do what you need to do with myInt
  } else {
    // what ever error code you need to write
  }

在Swift 4.2和Xcode 10.1中

let string = "789"
if let intValue = Int(string) {
    print(intValue)
}

let integerValue = 789
let stringValue = String(integerValue)

OR

let stringValue = "\(integerValue)"
print(stringValue)