I'm having a bit of issue with my code...right now, I am passing a string containing a bunch of numbers, to get converted to a number, comma separators added, then converted back to a string for output. When I add a decimal to my string and pass it in, a number like 996.3658 get truncated to 996.366...
"currentNumber" is my input value, "textOutputToScreen" is my output...
func formatNumber() {
let charset = CharacterSet(charactersIn: ".")
if let _ = currentNumber.rangeOfCharacter(from: charset) {
if let number = Float(currentNumber) {
let numberFormatter = NumberFormatter()
numberFormatter.numberStyle = .decimal
guard let formattedNumber = numberFormatter.string(from: NSNumber(value: number)) else { return }
textOutputToScreen = String(formattedNumber)
}
}
else {
if let number = Int(currentNumber) {
let numberFormatter = NumberFormatter()
numberFormatter.numberStyle = .decimal
guard let formattedNumber = numberFormatter.string(from: NSNumber(value: number)) else { return }
textOutputToScreen = String(formattedNumber)
}
}
}
Thank you in advance for your help!
The issue there is that you have to set your NumberFormatter minimumFractionDigits to 4. Btw there is no need to initialize a NSNumber object. You can use Formatters string(for: Any) method and pass your Float. Btw I would use a Double (64-bit) instead of a Float (32-bit) and there is no need to initialize a new string g from your formattedNumber object. It is already a String.
Another thing is that you don't need to know the location of the period you can simply use contains instead of rangeOfCharacter method. Your code should look something like this:
extension Formatter {
static let number: NumberFormatter = {
let formatter = NumberFormatter()
formatter.numberStyle = .decimal
return formatter
}()
}
func formatNumber(from string: String) -> String? {
if string.contains(".") {
guard let value = Double(string) else { return nil }
Formatter.number.minimumFractionDigits = 4
return Formatter.number.string(for: value)
} else {
guard let value = Int(string) else { return nil }
Formatter.number.minimumFractionDigits = 0
return Formatter.number.string(for: value)
}
}
let label = UILabel()
let currentNumber = "996.3658"
label.text = formatNumber(from: currentNumber) // "996.3658\n"
If you would like to assign the result to your var instead of a label
if let formatted = formatNumber(from: currentNumber) {
textOutputToScreen = formatted
}
I'm trying to work out how to cast an Int into a String in Swift.
I figure out a workaround, using NSNumber but I'd love to figure out how to do it all in Swift.
let x : Int = 45
let xNSNumber = x as NSNumber
let xString : String = xNSNumber.stringValue
Converting Int to String:
let x : Int = 42
var myString = String(x)
And the other way around - converting String to Int:
let myString : String = "42"
let x: Int? = myString.toInt()
if (x != nil) {
// Successfully converted String to Int
}
Or if you're using Swift 2 or 3:
let x: Int? = Int(myString)
Check the Below Answer:
let x : Int = 45
var stringValue = "\(x)"
print(stringValue)
Here are 4 methods:
var x = 34
var s = String(x)
var ss = "\(x)"
var sss = toString(x)
var ssss = x.description
I can imagine that some people will have an issue with ss. But if you were looking to build a string containing other content then why not.
In Swift 3.0:
var value: Int = 10
var string = String(describing: value)
Swift 4:
let x:Int = 45
let str:String = String(describing: x)
Developer.Apple.com > String > init(describing:)
The String(describing:) initializer is the preferred way to convert an instance of any type to a string.
Custom String Convertible
Just for completeness, you can also use:
let x = 10.description
or any other value that supports a description.
Swift 4:
Trying to show the value in label without Optional() word.
here x is a Int value using.
let str:String = String(x ?? 0)
To save yourself time and hassle in the future you can make an Int extension. Typically I create a shared code file where I put extensions, enums, and other fun stuff. Here is what the extension code looks like:
extension Int
{
func toString() -> String
{
var myString = String(self)
return myString
}
}
Then later when you want to convert an int to a string you can just do something like:
var myNumber = 0
var myNumberAsString = myNumber.toString()
in swift 3.0 this is how we can convert Int to String and String to Int
//convert Integer to String in Swift 3.0
let theIntegerValue :Int = 123 // this can be var also
let theStringValue :String = String(theIntegerValue)
//convert String to Integere in Swift 3.0
let stringValue : String = "123"
let integerValue : Int = Int(stringValue)!
for whatever reason the accepted answer did not work for me. I went with this approach:
var myInt:Int = 10
var myString:String = toString(myInt)
Multiple ways to do this :
var str1:String="\(23)"
var str2:String=String(format:"%d",234)
let intAsString = 45.description // "45"
let stringAsInt = Int("45") // 45
Swift 2:
var num1 = 4
var numString = "56"
var sum2 = String(num1) + numString
var sum3 = Int(numString)
Swift String performance
A little bit about performance
UI Testing Bundle on iPhone 7(real device) with iOS 14
let i = 0
lt result1 = String(i) //0.56s 5890kB
lt result2 = "\(i)" //0.624s 5900kB
lt result3 = i.description //0.758s 5890kB
import XCTest
class ConvertIntToStringTests: XCTestCase {
let count = 1_000_000
func measureFunction(_ block: () -> Void) {
let metrics: [XCTMetric] = [
XCTClockMetric(),
XCTMemoryMetric()
]
let measureOptions = XCTMeasureOptions.default
measureOptions.iterationCount = 5
measure(metrics: metrics, options: measureOptions) {
block()
}
}
func testIntToStringConstructor() {
var result = ""
measureFunction {
for i in 0...count {
result += String(i)
}
}
}
func testIntToStringInterpolation() {
var result = ""
measureFunction {
for i in 0...count {
result += "\(i)"
}
}
}
func testIntToStringDescription() {
var result = ""
measureFunction {
for i in 0...count {
result += i.description
}
}
}
}
iam using this simple approach
String to Int:
var a = Int()
var string1 = String("1")
a = string1.toInt()
and from Int to String:
var a = Int()
a = 1
var string1 = String()
string1= "\(a)"
Convert Unicode Int to String
For those who want to convert an Int to a Unicode string, you can do the following:
let myInteger: Int = 97
// convert Int to a valid UnicodeScalar
guard let myUnicodeScalar = UnicodeScalar(myInteger) else {
return ""
}
// convert UnicodeScalar to String
let myString = String(myUnicodeScalar)
// results
print(myString) // a
Or alternatively:
let myInteger: Int = 97
if let myUnicodeScalar = UnicodeScalar(myInteger) {
let myString = String(myUnicodeScalar)
}
I prefer using String Interpolation
let x = 45
let string = "\(x)"
Each object has some string representation. This makes things simpler. For example if you need to create some String with multiple values. You can also do any math in it or use some conditions
let text = "\(count) \(count > 1 ? "items" : "item") in the cart. Sum: $\(sum + shippingPrice)"
exampleLabel.text = String(yourInt)
To convert String into Int
var numberA = Int("10")
Print(numberA) // It will print 10
To covert Int into String
var numberA = 10
1st way)
print("numberA is \(numberA)") // It will print 10
2nd way)
var strSomeNumber = String(numberA)
or
var strSomeNumber = "\(numberA)"
let a =123456888
var str = String(a)
OR
var str = a as! String
In swift 3.0, you may change integer to string as given below
let a:String = String(stringInterpolationSegment: 15)
Another way is
let number: Int = 15
let _numberInStringFormate: String = String(number)
//or any integer number in place of 15
If you like swift extension, you can add following code
extension Int
{
var string:String {
get {
return String(self)
}
}
}
then, you can get string by the method you just added
var x = 1234
var s = x.string
let Str = "12"
let num: Int = 0
num = Int (str)
I am getting data from Json and displaying it in table view how to check whether the number is float or double or integer in swift 3 if it is float how to get the no.of digits after decimal can anyone help me how to implement this in swift 3 ?
if specialLoop.attributeCode == "special_price" {
let attributeString: NSMutableAttributedString = NSMutableAttributedString(string: "$ \((arr.price))")
attributeString.addAttribute(NSStrikethroughStyleAttributeName, value: 1, range: NSMakeRange(0, attributeString.length))
let specialPrice = specialLoop.value.replacingOccurrences(of: ".0000", with: "0")
print(specialPrice)
cell.productPrice.text = "$ \(specialPrice)"
cell.specialPriceLabel.isHidden = false
cell.specialPriceLabel.attributedText = attributeString
break
}
else {
cell.specialPriceLabel.isHidden = true
let price = arr.price
print(price)
cell.productPrice.text = "$ \( (price))0"
}
You can use (if let)
let data = [String: Any]()
if let value = data["key"] as? Int {
} else if let value = data["key"] as? Float {
} else if let value = data["key"] as? Double {
}
as describe below, you can find a type of any object (whether custom class or built-in class like - String, Int, etc.).
class demo {
let a: String = ""
}
let demoObj = demo()
print(type(of: demoObj))
--> Output: "demo.Type"
I am trying to convert a given double into scientific notation, and running into some problems. I cant seem to find much documentation on how to do it either. Currently I am using:
var val = 500
var numberFormatter = NSNumberFormatter()
numberFormatter.numberStyle = NSNumberFormatterStyle.ScientificStyle
let number = numberFormatter.numberFromString("\(val)")
println(number as Double?)
// Prints optional(500) instead of optional(5e+2)
What am I doing wrong?
You can set NumberFormatter properties positiveFormat and exponent Symbol to format your string as you want as follow:
let val = 500
let formatter = NumberFormatter()
formatter.numberStyle = .scientific
formatter.positiveFormat = "0.###E+0"
formatter.exponentSymbol = "e"
if let scientificFormatted = formatter.string(for: val) {
print(scientificFormatted) // "5e+2"
}
update: Xcode 9 • Swift 4
You can also create an extension to get a scientific formatted description from Numeric types as follow:
extension Formatter {
static let scientific: NumberFormatter = {
let formatter = NumberFormatter()
formatter.numberStyle = .scientific
formatter.positiveFormat = "0.###E+0"
formatter.exponentSymbol = "e"
return formatter
}()
}
extension Numeric {
var scientificFormatted: String {
return Formatter.scientific.string(for: self) ?? ""
}
}
print(500.scientificFormatted) // "5e+2"
The issue is that you are printing the number... not the formatted number. You are calling numberForString instead of stringForNumber
var val = 500
var numberFormatter = NSNumberFormatter()
numberFormatter.numberStyle = NSNumberFormatterStyle.ScientificStyle
let numberString = numberFormatter.stringFromNumber(val)
println(numberString)
Slight modification to the answer by leo-dabus to Xcode 9 Swift 4:
extension Double {
struct Number {
static var formatter = NumberFormatter()
}
var scientificStyle: String {
Number.formatter.numberStyle = .scientific
Number.formatter.positiveFormat = "0.###E+0"
Number.formatter.exponentSymbol = "e"
let number = NSNumber(value: self)
return Number.formatter.string(from :number) ?? description
}
}
I have used below code successfully to format Int with thousand separators. However my current project required Int64 to be formatted the same way and it throws error 'Int64' is not convertible to 'NSNumber'
var numberFormatter: NSNumberFormatter {
let formattedNumber = NSNumberFormatter()
formattedNumber.numberStyle = .DecimalStyle
formattedNumber.maximumFractionDigits = 0
return formattedNumber
}
You mean when you call numberFormatter.stringFromNumber(12345678) after the above code, like this?
let i64: Int64 = 1234567890
numberFormatter.stringFromNumber(i64)
Doesn’t look like Swift will cast from an Int64 to an NSNumber:
let i = 1234567890
let n = i as NSNumber // OK
numberFormatter.stringFromNumber(i) // Also OK
// Compiler error: 'Int64' is not convertible to 'NSNumber'
let n64 = i64 as NSNumber
// so the implicit conversion will also fail:
numberFormatter.stringFromNumber(i64)
This is a bit confounding, since Swift Ints are themselves usually the same size as Int64s.
You can work around it by constructing an NSNumber by hand:
let n64 = NSNumber(longLong: i64)
BTW beware that var trick: it’s nice that it encapsulates all the relevant code for creating numberFormatter, but that code will run afresh every time you use it. As an alternative you could do this:
let numberFormatter: NSNumberFormatter = {
let formattedNumber = NSNumberFormatter()
formattedNumber.numberStyle = .DecimalStyle
formattedNumber.maximumFractionDigits = 0
return formattedNumber
}()
If it’s a property in a struct/class, you could also make it a lazy var which has the added benefit of only being running if the variable is used, like your var, but only once.
struct Thing {
lazy var numberFormatter: NSNumberFormatter = {
println("blah")
let formattedNumber = NSNumberFormatter()
formattedNumber.numberStyle = .DecimalStyle
formattedNumber.maximumFractionDigits = 0
return formattedNumber
}()
}
extension Formatter {
static let decimalNumber: NumberFormatter = {
let formatter = NumberFormatter()
formatter.numberStyle = .decimal
return formatter
}()
}
extension Numeric {
var formatted: String { Formatter.decimalNumber.string(for: self) ?? "" }
}
let x: Int64 = 1000000
x.formatted // "1,000,000"