characters - an instance property of String, is deprecated from with Xcode 9.1
It was very useful to get a substring from String by using the characters property but now it has been deprecated and Xcode suggests to use substring. I've tried to check around SO questions and apple developer tutorials/guidelines for the same. But could not see any solution/alternate as suggested.
Here is warning message:
'characters' is deprecated: Please use String or Substring
I've so many string operations are performed/handled using property characters.
Anyone have any idea/info about this update?
Swift 4 introduced changes on string API.
You can just use !stringValue.isEmpty instead of stringValue.characters.count > 0
for more information you get the sample from here
for e.g
let edit = "Summary"
edit.count // 7
Swift 4 vs Swift 3 examples:
let myString = "test"
for char in myString.characters {print(char) } // Swift 3
for char in myString { print(char) } // Swift 4
let length = myString.characters.count // Swift 3
let length = myString.count // Swift 4
One of the most common cases for manipulating strings is with JSON responses. In this example I created an extension in my watch app to drop the last (n) characters of a Bitcoin JSON object.
Swift 3:
func dropLast(_ n: Int = 0) -> String {
return String(characters.dropLast(n))
Xcode 9.1 Error Message:
'characters' is deprecated: Please use String or Substring directly
Xcode is telling us to use the string variable or method directly.
Swift 4:
func dropLast(_ n: Int = 0) -> String {
return String(dropLast(n))
}
Complete Extension:
extension String {
func dropLast(_ n: Int = 0) -> String {
return String(dropLast(n))
}
var dropLast: String {
return dropLast()
}
}
Call:
print("rate:\(response.USDRate)")
let literalMarketPrice = response.USDRate.dropLast(2)
print("literal market price: \(literalMarketPrice)")
Console:
//rate:7,101.0888 //JSON float
//literal market price: 7,101.08 // JSON string literal
Additional Examples:
print("Spell has \(invisibleSpellName.count) characters.")
return String(dropLast(n))
return String(removeLast(n))
Documentation:
You'll often be using common methods such as dropLast() or removeLast() or count so here is the explicit Apple documentation for each method.
droplast()
removelast()
counting characters
Use this characters because String stopped being a collection in Swift 2.0. However this is still valid code in Swift 4 but is no longer necessary now that String is a Collection again.
For example a Swift 4 String now has a direct count property that gives the character count:
// Swift 4
let spString = "Stack"
spString.count // 5
Examples for String and SubString.
String
Swift 4 String now directly get Element that gives the first character of String: (string.characters.first)
let spString = "Stack"
let firstElement = spString.first //S
SubString
Using SubString get first character.
let spstring = "Welcome"
let indexStartOfText = spstring.index(spstring.startIndex, offsetBy: 1)
let sub = spstring.substring(to: indexStartOfText)
print(sub) //W
That warning is just a top of the iceberg, there were a loot of string changes, strings are again a collection of characters, but we got soemthing new and cool, subStrings :)
This is a great read about this:
https://useyourloaf.com/blog/updating-strings-for-swift-4/
Just remove characters
For example:
stringValue.characters.count
to
stringValue.count
You can also use this code for dictionary grouping without using { $0.characters.first! }.
let cities = ["Shanghai": 24_256_800, "Karachi": 23_500_000, "Beijing": 21_516_000, "Seoul": 9_995_000]
let groupedCities = Dictionary(grouping: cities.keys) { $0.first! }
print(groupedCities)
func validatePhoneNumber(number:String) -> Bool{
if number.count < 10. //deprecated ->(number.characters.count)
{
return false;
}else{
return true;
}
}
You use directly .count and characters is deprecated.
This question already has answers here:
How is optional binding used in swift?
(9 answers)
Closed 6 years ago.
I am working on a project which uses both swift an objective c. The team member before me have written this code in objective C ,which I am not familiar with. There is problem that most of the part involving storing and retrieving value from Sqlite is in obj C. This has been done in a common class to avoid Code redemption. However if i use swift to retrieve value through that obj C file a problem occur. If there is no value in that specified row it return "null".
Update: Checked for optional binding as said by Antony Raphel
Even if i check for nil directly before converting to 'as? String' the same error persist. I came to know that there is no equivalent of "null" in swift. Is there any hack to the value is empty (null) in swift?
Just replace your
var prevNotifCount = self.cmmn.getPreviousNotificationCount() as? String
and use
guard let prevNotifCount = self.cmmn.getPreviousNotificationCount() else{
print("No previous notification value")
return
}
no need to check for nil, if it will fail , else block will be executed
if let prevNotifCount = self.cmmn.getPreviousNotificationCount() as? String
{
self.cmmn.saveInDatabase("19", phoneNumber: "0", otp: "0")
print(self.cmmn.getPreviousNotificationCount())
}
else
{
print("No previous notification value")
}
This is standard Swift approach called optional binding. You safely unwrap an optional and if it is not nil assign it to a local variable
Try by adding if let to check nil condition like this:-
if let NotifCount = self.cmmn,getPreviousNotificationCount() as? String
{
prevNotifCount = NotifCount
}
Please try this, Hope it helps!
Use if let statement.
if let preNotifCount = self.cmmn.getPreviousNotofication {
//business logic
}
Now business logic would only be executed if preNotifCount is not nil.
Is there a way to print the runtime type of a variable in swift? For example:
var now = NSDate()
var soon = now.dateByAddingTimeInterval(5.0)
println("\(now.dynamicType)")
// Prints "(Metatype)"
println("\(now.dynamicType.description()")
// Prints "__NSDate" since objective-c Class objects have a "description" selector
println("\(soon.dynamicType.description()")
// Compile-time error since ImplicitlyUnwrappedOptional<NSDate> has no "description" method
In the example above, I'm looking for a way to show that the variable "soon" is of type ImplicitlyUnwrappedOptional<NSDate>, or at least NSDate!.
Update September 2016
Swift 3.0: Use type(of:), e.g. type(of: someThing) (since the dynamicType keyword has been removed)
Update October 2015:
I updated the examples below to the new Swift 2.0 syntax (e.g. println was replaced with print, toString() is now String()).
From the Xcode 6.3 release notes:
#nschum points out in the comments that the Xcode 6.3 release notes show another way:
Type values now print as the full demangled type name when used with
println or string interpolation.
import Foundation
class PureSwiftClass { }
var myvar0 = NSString() // Objective-C class
var myvar1 = PureSwiftClass()
var myvar2 = 42
var myvar3 = "Hans"
print( "String(myvar0.dynamicType) -> \(myvar0.dynamicType)")
print( "String(myvar1.dynamicType) -> \(myvar1.dynamicType)")
print( "String(myvar2.dynamicType) -> \(myvar2.dynamicType)")
print( "String(myvar3.dynamicType) -> \(myvar3.dynamicType)")
print( "String(Int.self) -> \(Int.self)")
print( "String((Int?).self -> \((Int?).self)")
print( "String(NSString.self) -> \(NSString.self)")
print( "String(Array<String>.self) -> \(Array<String>.self)")
Which outputs:
String(myvar0.dynamicType) -> __NSCFConstantString
String(myvar1.dynamicType) -> PureSwiftClass
String(myvar2.dynamicType) -> Int
String(myvar3.dynamicType) -> String
String(Int.self) -> Int
String((Int?).self -> Optional<Int>
String(NSString.self) -> NSString
String(Array<String>.self) -> Array<String>
Update for Xcode 6.3:
You can use the _stdlib_getDemangledTypeName():
print( "TypeName0 = \(_stdlib_getDemangledTypeName(myvar0))")
print( "TypeName1 = \(_stdlib_getDemangledTypeName(myvar1))")
print( "TypeName2 = \(_stdlib_getDemangledTypeName(myvar2))")
print( "TypeName3 = \(_stdlib_getDemangledTypeName(myvar3))")
and get this as output:
TypeName0 = NSString
TypeName1 = __lldb_expr_26.PureSwiftClass
TypeName2 = Swift.Int
TypeName3 = Swift.String
Original answer:
Prior to Xcode 6.3 _stdlib_getTypeName got the mangled type name of a variable. Ewan Swick's blog entry helps to decipher these strings:
e.g. _TtSi stands for Swift's internal Int type.
Mike Ash has a great blog entry covering the same topic.
Edit: A new toString function has been introduced in Swift 1.2 (Xcode 6.3).
You can now print the demangled type of any type using .self and any instance using .dynamicType:
struct Box<T> {}
toString("foo".dynamicType) // Swift.String
toString([1, 23, 456].dynamicType) // Swift.Array<Swift.Int>
toString((7 as NSNumber).dynamicType) // __NSCFNumber
toString((Bool?).self) // Swift.Optional<Swift.Bool>
toString(Box<SinkOf<Character>>.self) // __lldb_expr_1.Box<Swift.SinkOf<Swift.Character>>
toString(NSStream.self) // NSStream
Try calling YourClass.self and yourObject.dynamicType.
Reference: https://devforums.apple.com/thread/227425.
Swift 3.0
let string = "Hello"
let stringArray = ["one", "two"]
let dictionary = ["key": 2]
print(type(of: string)) // "String"
// Get type name as a string
String(describing: type(of: string)) // "String"
String(describing: type(of: stringArray)) // "Array<String>"
String(describing: type(of: dictionary)) // "Dictionary<String, Int>"
// Get full type as a string
String(reflecting: type(of: string)) // "Swift.String"
String(reflecting: type(of: stringArray)) // "Swift.Array<Swift.String>"
String(reflecting: type(of: dictionary)) // "Swift.Dictionary<Swift.String, Swift.Int>"
Is this what you're looking for?
println("\(object_getClassName(now))");
It prints "__NSDate"
UPDATE: Please note this no longer seems to work as of Beta05
My current Xcode is Version 6.0 (6A280e).
import Foundation
class Person { var name: String; init(name: String) { self.name = name }}
class Patient: Person {}
class Doctor: Person {}
var variables:[Any] = [
5,
7.5,
true,
"maple",
Person(name:"Sarah"),
Patient(name:"Pat"),
Doctor(name:"Sandy")
]
for variable in variables {
let typeLongName = _stdlib_getDemangledTypeName(variable)
let tokens = split(typeLongName, { $0 == "." })
if let typeName = tokens.last {
println("Variable \(variable) is of Type \(typeName).")
}
}
Output:
Variable 5 is of Type Int.
Variable 7.5 is of Type Double.
Variable true is of Type Bool.
Variable maple is of Type String.
Variable Swift001.Person is of Type Person.
Variable Swift001.Patient is of Type Patient.
Variable Swift001.Doctor is of Type Doctor.
As of Xcode 6.3 with Swift 1.2, you can simply convert type values into the full demangled String.
toString(Int) // "Swift.Int"
toString(Int.Type) // "Swift.Int.Type"
toString((10).dynamicType) // "Swift.Int"
println(Bool.self) // "Swift.Bool"
println([UTF8].self) // "Swift.Array<Swift.UTF8>"
println((Int, String).self) // "(Swift.Int, Swift.String)"
println((String?()).dynamicType)// "Swift.Optional<Swift.String>"
println(NSDate) // "NSDate"
println(NSDate.Type) // "NSDate.Type"
println(WKWebView) // "WKWebView"
toString(MyClass) // "[Module Name].MyClass"
toString(MyClass().dynamicType) // "[Module Name].MyClass"
You can still access the class, through className (which returns a String).
There are actually several ways to get the class, for example classForArchiver, classForCoder, classForKeyedArchiver (all return AnyClass!).
You can't get the type of a primitive (a primitive is not a class).
Example:
var ivar = [:]
ivar.className // __NSDictionaryI
var i = 1
i.className // error: 'Int' does not have a member named 'className'
If you want to get the type of a primitive, you have to use bridgeToObjectiveC(). Example:
var i = 1
i.bridgeToObjectiveC().className // __NSCFNumber
You can use reflect to get information about object.
For example name of object class:
var classname = reflect(now).summary
Xcode 8 Swift 3.0 use type(of:)
let className = "\(type(of: instance))"
I had luck with:
let className = NSStringFromClass(obj.dynamicType)
SWIFT 5
With the latest release of Swift 3 we can get pretty descriptions of type names through the String initializer. Like, for example print(String(describing: type(of: object))). Where object can be an instance variable like array, a dictionary, an Int, a NSDate, an instance of a custom class, etc.
Here is my complete answer: Get class name of object as string in Swift
That question is looking for a way to getting the class name of an object as string but, also i proposed another way to getting the class name of a variable that isn't subclass of NSObject. Here it is:
class Utility{
class func classNameAsString(obj: Any) -> String {
//prints more readable results for dictionaries, arrays, Int, etc
return String(describing: type(of: obj))
}
}
I made a static function which takes as parameter an object of type Any and returns its class name as String :) .
I tested this function with some variables like:
let diccionary: [String: CGFloat] = [:]
let array: [Int] = []
let numInt = 9
let numFloat: CGFloat = 3.0
let numDouble: Double = 1.0
let classOne = ClassOne()
let classTwo: ClassTwo? = ClassTwo()
let now = NSDate()
let lbl = UILabel()
and the output was:
diccionary is of type Dictionary
array is of type Array
numInt is of type Int
numFloat is of type CGFloat
numDouble is of type Double
classOne is of type: ClassOne
classTwo is of type: ClassTwo
now is of type: Date
lbl is of type: UILabel
In Xcode 8, Swift 3.0
Steps:
1. Get the Type:
Option 1:
let type : Type = MyClass.self //Determines Type from Class
Option 2:
let type : Type = type(of:self) //Determines Type from self
2. Convert Type to String:
let string : String = "\(type)" //String
In Swift 3.0, you can use type(of:), as dynamicType keyword has been removed.
To get a type of object or class of object in Swift, you must need to use a type(of: yourObject)
type(of: yourObject)
When using Cocoa (not CocoaTouch), you can use the className property for objects that are subclasses of NSObject.
println(now.className)
This property is not available for normal Swift objects, which aren't subclasses of NSObject (and in fact, there is no root id or object type in Swift).
class Person {
var name: String?
}
var p = Person()
println(person.className) // <- Compiler error
In CocoaTouch, at this time there is not a way to get a string description of the type of a given variable. Similar functionality also does not exist for primitive types in either Cocoa or CocoaTouch.
The Swift REPL is able to print out a summary of values including its type, so it is possible this manner of introspection will be possible via an API in the future.
EDIT: dump(object) seems to do the trick.
The top answer doesn't have a working example of the new way of doing this using type(of:. So to help rookies like me, here is a working example, taken mostly from Apple's docs here - https://developer.apple.com/documentation/swift/2885064-type
doubleNum = 30.1
func printInfo(_ value: Any) {
let varType = type(of: value)
print("'\(value)' of type '\(varType)'")
}
printInfo(doubleNum)
//'30.1' of type 'Double'
I've tried some of the other answers here but milage seems to very on what the underling object is.
However I did found a way you can get the Object-C class name for an object by doing the following:
now?.superclass as AnyObject! //replace now with the object you are trying to get the class name for
Here is and example of how you would use it:
let now = NSDate()
println("what is this = \(now?.superclass as AnyObject!)")
In this case it will print NSDate in the console.
I found this solution which hopefully might work for someone else.
I created a class method to access the value. Please bear in mind this will work for NSObject subclass only. But at least is a clean and tidy solution.
class var className: String!{
let classString : String = NSStringFromClass(self.classForCoder())
return classString.componentsSeparatedByString(".").last;
}
In the latest XCode 6.3 with Swift 1.2, this is the only way I found:
if view.classForCoder.description() == "UISegment" {
...
}
Many of the answers here do not work with the latest Swift (Xcode 7.1.1 at time of writing).
The current way of getting the information is to create a Mirror and interrogate that. For the classname it is as simple as:
let mirror = Mirror(reflecting: instanceToInspect)
let classname:String = mirror.description
Additional information about the object can also be retrieved from the Mirror. See http://swiftdoc.org/v2.1/type/Mirror/ for details.
Swift version 4:
print("\(type(of: self)) ,\(#function)")
// within a function of a class
Thanks #Joshua Dance
In lldb as of beta 5, you can see the class of an object with the command:
fr v -d r shipDate
which outputs something like:
(DBSalesOrderShipDate_DBSalesOrderShipDate_ *) shipDate = 0x7f859940
The command expanded out means something like:
Frame Variable (print a frame variable) -d run_target (expand dynamic types)
Something useful to know is that using "Frame Variable" to output variable values guarantees no code is executed.
I've found a solution for self-developed classes (or such you have access to).
Place the following computed property within your objects class definition:
var className: String? {
return __FILE__.lastPathComponent.stringByDeletingPathExtension
}
Now you can simply call the class name on your object like so:
myObject.className
Please note that this will only work if your class definition is made within a file that is named exactly like the class you want the name of.
As this is commonly the case the above answer should do it for most cases. But in some special cases you might need to figure out a different solution.
If you need the class name within the class (file) itself you can simply use this line:
let className = __FILE__.lastPathComponent.stringByDeletingPathExtension
Maybe this method helps some people out there.
Based on the answers and comments given by Klass and Kevin Ballard above, I would go with:
println(_stdlib_getDemangledTypeName(now).componentsSeparatedByString(".").last!)
println(_stdlib_getDemangledTypeName(soon).componentsSeparatedByString(".").last!)
println(_stdlib_getDemangledTypeName(soon?).componentsSeparatedByString(".").last!)
println(_stdlib_getDemangledTypeName(soon!).componentsSeparatedByString(".").last!)
println(_stdlib_getDemangledTypeName(myvar0).componentsSeparatedByString(".").last!)
println(_stdlib_getDemangledTypeName(myvar1).componentsSeparatedByString(".").last!)
println(_stdlib_getDemangledTypeName(myvar2).componentsSeparatedByString(".").last!)
println(_stdlib_getDemangledTypeName(myvar3).componentsSeparatedByString(".").last!)
which will print out:
"NSDate"
"ImplicitlyUnwrappedOptional"
"Optional"
"NSDate"
"NSString"
"PureSwiftClass"
"Int"
"Double"
let i: Int = 20
func getTypeName(v: Any) -> String {
let fullName = _stdlib_demangleName(_stdlib_getTypeName(i))
if let range = fullName.rangeOfString(".") {
return fullName.substringFromIndex(range.endIndex)
}
return fullName
}
println("Var type is \(getTypeName(i)) = \(i)")
Swift 4:
// "TypeName"
func stringType(of some: Any) -> String {
let string = (some is Any.Type) ? String(describing: some) : String(describing: type(of: some))
return string
}
// "ModuleName.TypeName"
func fullStringType(of some: Any) -> String {
let string = (some is Any.Type) ? String(reflecting: some) : String(reflecting: type(of: some))
return string
}
Usage:
print(stringType(of: SomeClass())) // "SomeClass"
print(stringType(of: SomeClass.self)) // "SomeClass"
print(stringType(of: String())) // "String"
print(fullStringType(of: String())) // "Swift.String"
There appears to be no generic way to print the type name of an arbitrary value's type. As others have noted, for class instances you can print value.className but for primitive values it appears that at runtime, the type information is gone.
For instance, it looks as if there's not a way to type: 1.something() and get out Int for any value of something. (You can, as another answer suggested, use i.bridgeToObjectiveC().className to give you a hint, but __NSCFNumber is not actually the type of i -- just what it will be converted to when it crosses the boundary of an Objective-C function call.)
I would be happy to be proven wrong, but it looks like the type checking is all done at compile time, and like C++ (with RTTI disabled) much of the type information is gone at runtime.
This is how you get a type string of your object or Type which is consistent and takes into account to which module the object definition belongs to or nested in. Works in Swift 4.x.
#inline(__always) func typeString(for _type: Any.Type) -> String {
return String(reflecting: type(of: _type))
}
#inline(__always) func typeString(for object: Any) -> String {
return String(reflecting: type(of: type(of: object)))
}
struct Lol {
struct Kek {}
}
// if you run this in playground the results will be something like
typeString(for: Lol.self) // __lldb_expr_74.Lol.Type
typeString(for: Lol()) // __lldb_expr_74.Lol.Type
typeString(for: Lol.Kek.self)// __lldb_expr_74.Lol.Kek.Type
typeString(for: Lol.Kek()) // __lldb_expr_74.Lol.Kek.Type
Not exactly what you are after, but you can also check the type of the variable against Swift types like so:
let object: AnyObject = 1
if object is Int {
}
else if object is String {
}
For example.
Xcode 7.3.1, Swift 2.2:
String(instanceToPrint.self).componentsSeparatedByString(".").last
I am trying to get to phone numbers of my contacts using CNContact, i want to have the number as a simple string of its didgits such as "04xxxxxxxx" but the closest I can get to is the following. ("contact" is of type CNContact)
contact.phoneNumbers[0].value
\\Which prints: <CNPhoneNumber: 0x13560a550: countryCode=au, digits=04xxxxxxxx>
ive tried all the obvious things and not so obvious things, thanks
If anyone has a more legitimate solution please do post it, otherwise this rather hacky approach works:
let value = String(contact.phoneNumbers[0].value)
let start = value.rangeOfString("digits=")!.endIndex
let end = value.endIndex.predecessor()
let number = value.substringWithRange(Range<String.Index>(start: start, end: end))
Using this category you can now access the phone number via swift.
don't forget to include this file in the bridging header
#implementation CNPhoneNumber (SwiftSupport)
- (NSString*)toString {
return self.stringValue;
}
#end
Fetch the number value as follows:
let number = value.valueForKey("digits") as! String
From iOS9:
let number = value.stringValue
According to Apple's documentation on CNPhoneNumber:
stringValue
Property
The string value of the phone number. (read-only)
Declaration
SWIFT
var stringValue: String { get }
Just downloaded Xcode 7 Beta, and this error appeared on enumerate keyword.
for (index, string) in enumerate(mySwiftStringArray)
{
}
Can anyone help me overcome this ?
Also, seems like count() is no longer working for counting length of String.
let stringLength = count(myString)
On above line, compiler says :
'count' is unavailable: access the 'count' property on the collection.
Has Apple has released any programming guide for Swift 2.0 ?
Many global functions have been replaced by protocol extension methods,
a new feature of Swift 2, so enumerate() is now an extension method
for SequenceType:
extension SequenceType {
func enumerate() -> EnumerateSequence<Self>
}
and used as
let mySwiftStringArray = [ "foo", "bar" ]
for (index, string) in mySwiftStringArray.enumerate() {
print(string)
}
And String does no longer conform to SequenceType, you have to
use the characters property to get the collection of Unicode
characters. Also, count() is a protocol extension method of
CollectionType instead of a global function:
let myString = "foo"
let stringLength = myString.characters.count
print(stringLength)
Update for Swift 3: enumerate() has been renamed to enumerated():
let mySwiftStringArray = [ "foo", "bar" ]
for (index, string) in mySwiftStringArray.enumerated() {
print(string)
}
There was an update for Swift 2 on using enumerate().
Instead of enumerate(...), people should use
... .enumerate()
The reason is that many global functions have been replaced by protocol extension methods and they will get an enumerate error.
Hope this helps.
All the best.
n
I know this is a old thread but I've just been messing around with Swift 2.0 and Playgrounds and I came across the same problem I thought I'd share a solution which uses the enumerate() method for a String
// This line works in Swift 1.2
// for (idx, character) in enumerate("A random string, it has a comma.")
// Swift 2.x
let count = inputString.characters
for (idx, character) in count.enumerate() where character == "," {
// Do something with idx
}
Hope this helps
Thanks
Kai