至于swift 3，我必须强制我的#%@!字符串和int加上一个“!”否则它就不起作用。

例如:

let prefs = UserDefaults.standard
var counter: String!
counter = prefs.string(forKey:"counter")
print("counter: \(counter!)")


var counterInt = Int(counter!)
counterInt = counterInt! + 1
print("counterInt: \(counterInt!)")

OUTPUT:
counter: 1
counterInt: 2

//  To convert user input (i.e string) to int for calculation.I did this , and it works.


    let num:Int? = Int(firstTextField.text!);

    let sum:Int = num!-2

    print(sum);

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

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
  }

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3.0

试试这个，你不需要检查任何条件，我已经做了一切，只是使用这个函数。发送任何字符串，数字，浮点数，双精度等。您将得到一个数字作为值，如果它无法转换您的值则为0

功能:

func getNumber(number: Any?) -> NSNumber {
    guard let statusNumber:NSNumber = number as? NSNumber else
    {
        guard let statString:String = number as? String else
        {
            return 0
        }
        if let myInteger = Int(statString)
        {
            return NSNumber(value:myInteger)
        }
        else{
            return 0
        }
    }
    return statusNumber
}

用法: 在代码中添加上述函数并转换使用 让myNumber = getNumber(number: myString) 如果myString包含数字或字符串，则返回数字，否则返回0

示例1:

let number:String = "9834"
print("printing number \(getNumber(number: number))")

输出:打印号9834

示例2:

let number:Double = 9834
print("printing number \(getNumber(number: number))")

输出:打印号9834

示例3:

let number = 9834
print("printing number \(getNumber(number: number))")

输出:打印号9834