在 Swift 中将字符串转换为 Float

Double() builds an Double from an Int, like this:

var convertedDouble = Double(someInt)

Note that this will only work if your text actually contains a number. Since Wage is a text field, the user can enter whatever they want and this will trigger a runtime error when you go to unbox the Optional returned from toInt(). You should check that the conversion succeeded before forcing the unboxing.

if let wageInt = Wage.text?.toInt() {
//we made it in the if so the conversion succeeded.
var wageConversionDouble = Double(wageInt)
}

Edit:

If you're sure the text will be an integer, you can do something like this (note that text on UITextField is also Optional)):

if let wageText = Wage.text {
var wageFloat = Double(wageText.toInt()!)
}

Swift 2.0+

Now with Swift 2.0 you can just use Float(Wage.text) which returns a Float? type. More clear than the below solution which just returns 0.

If you want a 0 value for an invalid Float for some reason you can use Float(Wage.text) ?? 0 which will return 0 if it is not a valid Float.

Old Solution

The best way to handle this is direct casting:

var WageConversion = (Wage.text as NSString).floatValue

I actually created an extension to better use this too:

extension String {
var floatValue: Float {
return (self as NSString).floatValue
}
}

Now you can just call var WageConversion = Wage.text.floatValue and allow the extension to handle the bridge for you!

This is a good implementation since it can handle actual floats (input with .) and will also help prevent the user from copying text into your input field (12p.34, or even 12.12.41).

Obviously, if Apple does add a floatValue to Swift this will throw an exception, but it may be nice in the mean time. If they do add it later, then all you need to do to modify your code is remove the extension and everything will work seamlessly, since you will already be calling .floatValue!

Also, variables and constants should start with a lower case (including IBOutlets)

Easy way:

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


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

you can use same approach for other calculations, hope this help !!

Use this:

 // get the values from text boxes
let a:Double = firstText.text.bridgeToObjectiveC().doubleValue
let b:Double = secondText.text.bridgeToObjectiveC().doubleValue


//  we checking against 0.0, because above function return 0.0 if it gets failed to convert
if (a != 0.0) && (b != 0.0) {
var ans = a + b
answerLabel.text = "Answer is \(ans)"
} else {
answerLabel.text = "Input values are not numberic"
}

You have two options which are quite similar (by the approach and result):

// option 1:
var string_1 : String = "100"
var double_1 : Double = (string_1 as NSString).doubleValue + 99.0


// option 2:
var string_2 : NSString = "100"
// or:  var string_2 = "100" as NSString
var number_2 : Double = string_2.doubleValue;

Because in some parts of the world, for example, a comma is used instead of a decimal. It is best to create a NSNumberFormatter to convert a string to float.

let numberFormatter = NSNumberFormatter()
numberFormatter.numberStyle = NSNumberFormatterStyle.DecimalStyle
let number = numberFormatter.numberFromString(self.stringByTrimmingCharactersInSet(Wage.text))

import Foundation
"-23.67".floatValue // => -23.67


let s = "-23.67" as NSString
s.floatValue // => -23.67

This is how I approached it. I did not want to "cross the bridge", as it has been removed from Xcode 6 beta 5 anyway, quick and dirty:

extension String {


// converting a string to double
func toDouble() -> Double? {


// split the string into components
var comps = self.componentsSeparatedByString(".")


// we have nothing
if comps.count == 0 {
return nil
}
// if there is more than one decimal
else if comps.count > 2 {
return nil
}
else if comps[0] == "" || comps[1] == "" {
return nil
}


// grab the whole portion
var whole = 0.0
// ensure we have a number for the whole
if let w = comps[0].toInt() {
whole = Double(w)
}
else {
return nil
}


// we only got the whole
if comps.count == 1 {


return whole


}


// grab the fractional
var fractional = 0.0
// ensure we have a number for the fractional
if let f = comps[1].toInt() {


// use number of digits to get the power
var toThePower = Double(countElements(comps[1]))


// compute the fractional portion
fractional = Double(f) / pow(10.0, toThePower)


}
else {
return nil
}


// return the result
return whole + fractional
}


// converting a string to float
func toFloat() -> Float? {


if let val = self.toDouble() {
return Float(val)
}
else {
return nil
}


}


}




// test it out
var str = "78.001"


if let val = str.toFloat() {
println("Str in float: \(val)")
}
else {
println("Unable to convert Str to float")
}


// now in double
if let val = str.toDouble() {
println("Str in double: \(val)")
}
else {
println("Unable to convert Str to double")
}

I convert String to Float in this way:

let numberFormatter = NSNumberFormatter()
let number = numberFormatter.numberFromString("15.5")
let numberFloatValue = number.floatValue


println("number is \(numberFloatValue)") // prints "number is 15.5"

Using the accepted solution, I was finding that my "1.1" (when using the .floatValue conversion) would get converted to 1.10000002384186, which was not what I wanted. However, if I used the .doubleValue instead, I would get the 1.1 that I wanted.

So for example, instead of using the accepted solution, I used this instead:

var WageConversion = (Wage.text as NSString).doubleValue

In my case I did not need double-precision, but using the .floatValue was not giving me the proper result.

Just wanted to add this to the discussion in case someone else had been running into the same issue.

Update

The accepted answer shows a more up to date way of doing

Swift 1

This is how Paul Hegarty has shown on Stanford's CS193p class in 2015:

wageConversion = NSNumberFormatter().numberFromString(wage.text!)!.floatValue

You can even create a computed property for not having to do that every time

var wageValue: Float {
get {
return NSNumberFormatter().numberFromString(wage.text!)!.floatValue
}
set {
wage.text = "\(newValue)"
}
}

Below will give you an optional Float, stick a ! at the end if you know it to be a Float, or use if/let.

let wageConversion = Float(wage.text)

extension String {
func floatValue() -> Float? {
return Float(self)
}
}

Here is a Swift 3 adaptation of Paul Hegarty's solution from rdprado's answer, with some checking for optionals added to it (returning 0.0 if any part of the process fails):

var wageFloat:Float = 0.0


if let wageText = wage.text {
if let wageNumber = NumberFormatter().number(from: wageText) {
wageFloat = wageNumber.floatValue
}
}

By the way, I took Stanford's CS193p class using iTunes University when it was still teaching Objective-C.

I found Paul Hegarty to be a FANTASTIC instructor, and I would highly recommend the class to anyone starting out as an iOS developer in Swift!!!

For the sake of completeness this is a solution using an extension of UITextField which can also consider a different locale.

For Swift 3+

extension UITextField {
func floatValue(locale : Locale = Locale.current) -> Float {
let numberFormatter = NumberFormatter()
numberFormatter.numberStyle = .decimal
numberFormatter.locale = locale


let nsNumber = numberFormatter.number(from: text!)
return nsNumber == nil ? 0.0 : nsNumber!.floatValue
}
}

I found another way to take a input value of a UITextField and cast it to a float:

    var tempString:String?
var myFloat:Float?


@IBAction func ButtonWasClicked(_ sender: Any) {
tempString = myUITextField.text
myFloat = Float(tempString!)!
}

In swift 4

let Totalname = "10.0" //Now it is in string


let floatVal  = (Totalname as NSString).floatValue //Now converted to float

Swift 4/5, use just Float(value).

let string_value : String = "123200"


let float_value : Float = Float(string_value)


print(float_value)

Answer: 123200

you can use,

let wg = Float(wage.text!)

If you want to round the float to 2 decimal places:

let wg = Float(String(format: "%.2f",wage.text!)

Works on Swift 5+


import Foundation


let myString:String = "50"
let temp = myString as NSString
let myFloat = temp.floatValue
print(myFloat)  //50.0
print(type(of: myFloat)) // Float

// Also you can guard your value in order to check what is happening whenever your app crashes.

guard let myFloat = temp.floatValue else {
fatalError(" fail to change string to float value.")
}

to convert string to Float in Xcode 11 as previous methods need modification

func stringToFloat(value : String) -> Float {
let numberFormatter = NumberFormatter()
let number = numberFormatter.number(from: value)
let numberFloatValue = number?.floatValue
return numberFloatValue!
}

I convert String to Float in this way:

let wage:Float = Float(textField.text!)!