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?

try this NSNumberFormatter().numberFromString("42")!.doubleValue — TonyMkenu, Feb 19 '15 at 14:27

score 161 · Accepted Answer · edited Jun 20 '20 at 09:12

161

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")
}

edited Jun 20 '20 at 09:12

Community

1
1

answered Feb 19 '15 at 11:55

Wez

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my number is long long not interger and replacing interger to longlong shows error? – user2413621 Feb 19 '15 at 12:17
are you saying that your string is not in this format: "42" ? – Wez Feb 19 '15 at 12:23
1

Perhaps you could update your question with a string in the format you want? – Wez Feb 19 '15 at 13:09
Swap 2.0 and 1.0 answers, placing 2.0 on top. – SwiftArchitect Oct 25 '15 at 22:38
@SwiftArchitect feel free to edit my answer if you feel it could be structured better. – Wez Oct 25 '15 at 22:45
If I do it will probably be declined as "no improvement" (-: – SwiftArchitect Oct 25 '15 at 23:20
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 – emem Jun 30 '16 at 12:41
They couldn't bang our asses any other way, so they made swift... Putting us in a position to ask such simple questions. – Chanchal Raj Nov 03 '16 at 07:34
Keep in mind, the string may contain double value, not int. In this case you should use if let myInteger = Double(someString) – Aleksey Tsyss Apr 10 '18 at 11:17

score 16 · Answer 2 · answered Jul 21 '15 at 18:32

16

Or do it just in one line:

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

answered Jul 21 '15 at 18:32

polarware

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3

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. – Harry Bloom Dec 21 '17 at 13:13
1

Has been updated for Swift 5: ```NumberFormatter().number(from: "55")!``` – Anna Billstrom Nov 14 '20 at 23:21

score 9 · Answer 3 · answered Sep 04 '19 at 08:02

9

In latest Swift:

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

answered Sep 04 '19 at 08:02

Hemang

26,840
19
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186

score 8 · Answer 4 · edited Feb 03 '21 at 19:52

8

Swift 2

Try this:

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

Swift 3.x

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

edited Feb 03 '21 at 19:52

Noor

967
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18

answered Aug 04 '16 at 11:09

Vikram Biwal

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1

with Swift 3.x this is now.. `NSNumber.init( value: Int32("22")!)` – Athir Nuaimi Aug 25 '16 at 01:41
2

no need to initialize with '.init'. simply write: NSNumber(value: Int32("22")!) – Peter Kreinz Apr 25 '17 at 13:32

rakeshbs · Answer 5 · 2015-02-20T10:12:45.030

6

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)
}

edited Feb 20 '15 at 10:12

answered Feb 19 '15 at 11:59

rakeshbs

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Note also that `NSNumberFormatter` is expensive to create, and should be cached whenever possible instead of being created every time it's used. – David Berry Feb 19 '15 at 14:28

score 2 · Answer 6 · answered Apr 18 '18 at 01:14

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)

score 2 · Answer 7 · answered Jun 01 '21 at 07:06

2

Swift 5

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

answered Jun 01 '21 at 07:06

Manish

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Ahmed Samir · Answer 8 · 2020-04-16T03:39:13.003

-1

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

and so on .

edited Apr 16 '20 at 03:39

answered Apr 16 '20 at 03:00

Ahmed Samir

291
3
6

Rameshios · Answer 9 · 2021-04-29T15:51:52.450

-5

Try Once

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

edited Apr 29 '21 at 15:51

answered Oct 18 '16 at 12:34

Rameshios

1
4

String to NSNumber in Swift

9 Answers9

Swift 3.0

Swift 2.0

Swift 1.0