String to NSNumber in Swift

Question

I found a method to convert String to NSNumber, but the code is in Objective-C. I have tried converting it to Swift but it is not working.

The code I am using:

NSNumberFormatter *f = [[NSNumberFormatter alloc] init];
f.numberStyle = NSNumberFormatterDecimalStyle;
NSNumber *myNumber = [f numberFromString:@"42222222222"];

and in Swift I am using it in this way:

var i = NSNumberFormatter.numberFromString("42")

But this code is not working. What am I doing wrong?

Vansickle

try this NSNumberFormatter().numberFromString("42")!.doubleValue – Unblock

Answer 1

Swift 3.0

NSNumber(integer:myInteger) has changed to NSNumber(value:myInteger)

let someString = "42222222222"
if let myInteger = Int(someString) {
    let myNumber = NSNumber(value:myInteger)
}

---

Swift 2.0

Use the Int() initialiser like this.

let someString = "42222222222"
if let myInteger = Int(someString) {
    let myNumber = NSNumber(integer:myInteger)
    print(myNumber)
} else {
    print("'\(someString)' did not convert to an Int")
}

This can be done in one line if you already know the string will convert perfectly or you just don't care.

let myNumber = Int("42222222222")!

---

Swift 1.0

Use the toInt() method.

let someString = "42222222222"
if let myInteger = someString.toInt() {
    let myNumber = NSNumber(integer:myInteger)
    println(myNumber)
} else {
    println("'\(someString)' did not convert to an Int")
}

Hepta

my number is long long not interger and replacing interger to longlong shows error? – Vansickle

are you saying that your string is not in this format: "42" ? – Hepta

Perhaps you could update your question with a string in the format you want? – Hepta

Swap 2.0 and 1.0 answers, placing 2.0 on top. – Castrate

@Castrate feel free to edit my answer if you feel it could be structured better. – Hepta

If I do it will probably be declined as "no improvement" (-: – Castrate

You should be really careful with using Int and Int64 - if the number can be bigger than Int32 and your app can run on iPhone 5 or older, use Int64 and not Int – Zoochore

They couldn't bang our asses any other way, so they made swift... Putting us in a position to ask such simple questions. – Semi

Keep in mind, the string may contain double value, not int. In this case you should use if let myInteger = Double(someString) – Niobium

Answer 2

Or do it just in one line:

NSNumberFormatter().numberFromString("55")!.decimalValue

Gertiegertrud

Like David Berry commented below, be careful with creating NSNumberFormatters like this as they are expensive to create. Better caching or using a singleton if using the same formatter more than once. – Electrotherapeutics

Has been updated for Swift 5: NumberFormatter().number(from: "55")! – Rigid

Answer 3

In latest Swift:

let number = NumberFormatter().number(from: "1234")

Aleksandropol

Answer 4

Swift 2

Try this:

var num = NSNumber(int: Int32("22")!)

Swift 3.x

 NSNumber(value: Int32("22")!)

Clasp

with Swift 3.x this is now.. NSNumber.init( value: Int32("22")!) – Tapley

no need to initialize with '.init'. simply write: NSNumber(value: Int32("22")!) – Osmunda

Answer 5

You can use the following code if you must use NSNumberFormatter. It's simpler to use Wezly's method.

let formatter = NSNumberFormatter()
formatter.numberStyle = NSNumberFormatterStyle.DecimalStyle;
if let number = formatter.numberFromString("42") {
    println(number)
}

Wreckage

Note also that NSNumberFormatter is expensive to create, and should be cached whenever possible instead of being created every time it's used. – Proven

Answer 6

Swift 5

let myInt = NumberFormatter().number(from: "42")

Twila

Answer 7

I do use extension in swift 3/4 and it's cool.

extension String {
    var numberValue: NSNumber? {
        if let value = Int(self) {
            return NSNumber(value: value)
        }
        return nil
    }
}

and then just use following code:

stringVariable.numberValue

What is cool is that you don't need a chain of if statements to unwrap the optional values. For instance,

if let _ = stringVariable, let intValue = Int(stringVariable!) {
    doSomething(NSNumber.init(value: intValue))
}

can be replaced by:

doSomething(stringVariable?.numberValue)

Batman

Answer 8

("23" as NSString).integerValue ("23.5" as NSString).doubleValue

and so on .

Punch

Answer 9

Try Once

let myString = "123"
let myInt = NSNumber(value: Int(myString) ?? 0)

Droppings