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I know how to programmatically do it, but I'm sure there's a built-in way...
Every language I've used has some sort of default textual representation for a collection of objects that it will spit out when you try to concatenate the Array with a string, or pass it to a print() function, etc. Does Apple's Swift language have a built-in way of easily turning an Array into a String, or do we always have to be explicit when stringifying an array?
If the array contains strings, you can use the String's join method:
var array = ["1", "2", "3"]
let stringRepresentation = "-".join(array) // "1-2-3"
In Swift 2:
var array = ["1", "2", "3"]
let stringRepresentation = array.joinWithSeparator("-") // "1-2-3"
This can be useful if you want to use a specific separator (hypen, blank, comma, etc).
Otherwise you can simply use the description property, which returns a string representation of the array:
let stringRepresentation = [1, 2, 3].description // "[1, 2, 3]"
Hint: any object implementing the Printable protocol has a description property. If you adopt that protocol in your own classes/structs, you make them print friendly as well
In Swift 3
join becomes joined, example [nil, "1", "2"].flatMap({$0}).joined()
joinWithSeparator becomes joined(separator:) (only available to Array of Strings)
In Swift 4
var array = ["1", "2", "3"]
array.joined(separator:"-")
With Swift 5, according to your needs, you may choose one of the following Playground sample codes in order to solve your problem.
Turning an array of Characters into a String with no separator:
let characterArray: [Character] = ["J", "o", "h", "n"]
let string = String(characterArray)
print(string)
// prints "John"
Turning an array of Strings into a String with no separator:
let stringArray = ["Bob", "Dan", "Bryan"]
let string = stringArray.joined(separator: "")
print(string) // prints: "BobDanBryan"
Turning an array of Strings into a String with a separator between words:
let stringArray = ["Bob", "Dan", "Bryan"]
let string = stringArray.joined(separator: " ")
print(string) // prints: "Bob Dan Bryan"
Turning an array of Strings into a String with a separator between characters:
let stringArray = ["car", "bike", "boat"]
let characterArray = stringArray.flatMap { $0 }
let stringArray2 = characterArray.map { String($0) }
let string = stringArray2.joined(separator: ", ")
print(string) // prints: "c, a, r, b, i, k, e, b, o, a, t"
Turning an array of Floats into a String with a separator between numbers:
let floatArray = [12, 14.6, 35]
let stringArray = floatArray.map { String($0) }
let string = stringArray.joined(separator: "-")
print(string)
// prints "12.0-14.6-35.0"
Swift 2.0 Xcode 7.0 beta 6 onwards uses joinWithSeparator() instead of join():
var array = ["1", "2", "3"]
let stringRepresentation = array.joinWithSeparator("-") // "1-2-3"
joinWithSeparator is defined as an extension on SequenceType
extension SequenceType where Generator.Element == String {
/// Interpose the `separator` between elements of `self`, then concatenate
/// the result. For example:
///
/// ["foo", "bar", "baz"].joinWithSeparator("-|-") // "foo-|-bar-|-baz"
#warn_unused_result
public func joinWithSeparator(separator: String) -> String
}
Swift 3
["I Love","Swift"].joined(separator:" ") // previously joinWithSeparator(" ")
Since no one has mentioned reduce, here it is:
[0, 1, 1, 0].map {"\($0)"}.reduce("") { $0 + $1 } // "0110"
In the spirit of functional programming 🤖
In Swift 4
let array:[String] = ["Apple", "Pear ","Orange"]
array.joined(separator: " ")
To change an array of Optional/Non-Optional Strings
//Array of optional Strings
let array : [String?] = ["1",nil,"2","3","4"]
//Separator String
let separator = ","
//flatMap skips the nil values and then joined combines the non nil elements with the separator
let joinedString = array.flatMap{ $0 }.joined(separator: separator)
//Use Compact map in case of **Swift 4**
let joinedString = array.compactMap{ $0 }.joined(separator: separator
print(joinedString)
Here flatMap, compactMap skips the nil values in the array and appends the other values to give a joined string.
Nowadays, in iOS 13+ and macOS 10.15+, we might use ListFormatter:
let formatter = ListFormatter()
let names = ["Moe", "Larry", "Curly"]
if let string = formatter.string(from: names) {
print(string)
}
That will produce a nice, natural language string representation of the list. A US user will see:
Moe, Larry, and Curly
It will support any languages for which (a) your app has been localized; and (b) the user’s device is configured. For example, a German user with an app supporting German localization, would see:
Moe, Larry und Curly
Mine works on NSMutableArray with componentsJoinedByString
var array = ["1", "2", "3"]
let stringRepresentation = array.componentsJoinedByString("-") // "1-2-3"
In Swift 2.2 you may have to cast your array to NSArray to use componentsJoinedByString(",")
let stringWithCommas = (yourArray as NSArray).componentsJoinedByString(",")
let arrayTemp :[String] = ["Mani","Singh","iOS Developer"]
let stringAfterCombining = arrayTemp.componentsJoinedByString(" ")
print("Result will be >>> \(stringAfterCombining)")
Result will be >>> Mani Singh iOS Developer
If you want to ditch empty strings in the array.
["Jet", "Fire"].filter { !$0.isEmpty }.joined(separator: "-")
If you want to filter nil values as well:
["Jet", nil, "", "Fire"].flatMap { $0 }.filter { !$0.isEmpty }.joined(separator: "-")
you can use joined() to get a single String when you have array of struct also.
struct Person{
let name:String
let contact:String
}
You can easily produce string using map() & joined()
PersonList.map({"\($0.name) - \($0.contact)"}).joined(separator: " | ")
output:
Jhon - 123 | Mark - 456 | Ben - 789
if you want convert custom object array to string or comma separated string (csv) you can use
var stringIds = (self.mylist.map{$0.id ?? 0}).map{String($0)}.joined(separator: ",")
credit to : urvish modi
post: Convert an array of Ints to a comma separated string
A separator can be a bad idea for some languages like Hebrew or Japanese.
Try this:
// Array of Strings
let array: [String] = ["red", "green", "blue"]
let arrayAsString: String = array.description
let stringAsData = arrayAsString.data(using: String.Encoding.utf16)
let arrayBack: [String] = try! JSONDecoder().decode([String].self, from: stringAsData!)
For other data types respectively:
// Set of Doubles
let set: Set<Double> = [1, 2.0, 3]
let setAsString: String = set.description
let setStringAsData = setAsString.data(using: String.Encoding.utf16)
let setBack: Set<Double> = try! JSONDecoder().decode(Set<Double>.self, from: setStringAsData!)
You can either use loops for getting this done. Or by using map.
By mapping:
let array = ["one" , "two" , "three"]
array.map({$0}).joined(seperator : ",")
so in separator you can modify the string.
Output-> ("one,two,three")
The Swift equivalent to what you're describing is string interpolation. If you're thinking about things like JavaScript doing "x" + array, the equivalent in Swift is "x\(array)".
As a general note, there is an important difference between string interpolation vs the Printable protocol. Only certain classes conform to Printable. Every class can be string interpolated somehow. That's helpful when writing generic functions. You don't have to limit yourself to Printable classes.
You can print any object using the print function
or use \(name) to convert any object to a string.
Example:
let array = [1,2,3,4]
print(array) // prints "[1,2,3,4]"
let string = "\(array)" // string == "[1,2,3,4]"
print(string) // prints "[1,2,3,4]"
Create extension for an Array:
extension Array {
var string: String? {
do {
let data = try JSONSerialization.data(withJSONObject: self, options: [.prettyPrinted])
return String(data: data, encoding: .utf8)
} catch {
return nil
}
}
}
if you have string array list , then convert to Int
let arrayList = list.map { Int($0)!}
arrayList.description
it will give you string value
for any Element type
extension Array {
func joined(glue:()->Element)->[Element]{
var result:[Element] = [];
result.reserveCapacity(count * 2);
let last = count - 1;
for (ix,item) in enumerated() {
result.append(item);
guard ix < last else{ continue }
result.append(glue());
}
return result;
}
}
Try This:
let categories = dictData?.value(forKeyPath: "listing_subcategories_id") as! NSMutableArray
let tempArray = NSMutableArray()
for dc in categories
{
let dictD = dc as? NSMutableDictionary
tempArray.add(dictD?.object(forKey: "subcategories_name") as! String)
}
let joinedString = tempArray.componentsJoined(by: ",")
use this when you want to convert list of struct type into string
struct MyStruct {
var name : String
var content : String
}
let myStructList = [MyStruct(name: "name1" , content: "content1") , MyStruct(name: "name2" , content: "content2")]
and covert your array like this way
let myString = myStructList.map({$0.name}).joined(separator: ",")
will produce ===> "name1,name2"
FOR SWIFT 3:
func textField(textField: UITextField, shouldChangeCharactersInRange range: NSRange, replacementString string: String) -> Bool {
if textField == phoneField
{
let newString = NSString(string: textField.text!).replacingCharacters(in: range, with: string)
let components = newString.components(separatedBy: NSCharacterSet.decimalDigits.inverted)
let decimalString = NSString(string: components.joined(separator: ""))
let length = decimalString.length
let hasLeadingOne = length > 0 && decimalString.character(at: 0) == (1 as unichar)
if length == 0 || (length > 10 && !hasLeadingOne) || length > 11
{
let newLength = NSString(string: textField.text!).length + (string as NSString).length - range.length as Int
return (newLength > 10) ? false : true
}
var index = 0 as Int
let formattedString = NSMutableString()
if hasLeadingOne
{
formattedString.append("1 ")
index += 1
}
if (length - index) > 3
{
let areaCode = decimalString.substring(with: NSMakeRange(index, 3))
formattedString.appendFormat("(%#)", areaCode)
index += 3
}
if length - index > 3
{
let prefix = decimalString.substring(with: NSMakeRange(index, 3))
formattedString.appendFormat("%#-", prefix)
index += 3
}
let remainder = decimalString.substring(from: index)
formattedString.append(remainder)
textField.text = formattedString as String
return false
}
else
{
return true
}
}
If you question is something like this:
tobeFormattedString = ["a", "b", "c"]
Output = "abc"
String(tobeFormattedString)
subscript (r: Range<Int>) -> String {
let start = startIndex.advancedBy(r.startIndex)
let end = start.advancedBy(r.endIndex - r.startIndex)
return self[Range(start: start, end: end)]
}
Struggling to convert the above subscript in my String extension to swift 3. Below is what happened after I pressed the convert button on Xcode.
subscript (r: Range<Int>) -> String {
let start = characters.index(startIndex, offsetBy: r.lowerBound)
let end = <#T##String.CharacterView corresponding to `start`##String.CharacterView#>.index(start, offsetBy: r.upperBound - r.lowerBound)
return self[(start ..< end)]
}
Screenshot of error
All you need to do is to add characters in front of index. The complier also gives you a hint to add a String.CharacterView corresponding tostart##String.CharacterView. The message is maybe a bit fuzzy, but it contains great value! Tells you, that is expecting an array of characters. However, as #vadian suggests, you can even omit the characters from the beginning.
I have written a little test as well, just to make sure.
import Foundation
extension String {
subscript (r: Range<Int>) -> String {
let start = index(startIndex, offsetBy: r.lowerBound)
let end = index(start, offsetBy: r.upperBound - r.lowerBound)
return self[start..<end]
}
}
let string = "Hello world"
let range = Range(uncheckedBounds: (lower: 0, upper: 2))
let s = string[range] // prints "He"
What is the most efficient way to remove all the spaces, \n and \r in a String in Swift?
I have tried:
for character in string.characters {
}
But it's a little inconvenient.
Swift 4:
let text = "This \n is a st\tri\rng"
let test = String(text.filter { !" \n\t\r".contains($0) })
Output:
print(test) // Thisisastring
While Fahri's answer is nice, I prefer it to be pure Swift ;)
edit/update:
Swift 5.2 or later
We can use the new Character property isWhitespace
let textInput = "Line 1 \n Line 2 \n\r"
let result = textInput.filter { !$0.isWhitespace }
result // "Line1Line2"
extension StringProtocol where Self: RangeReplaceableCollection {
var removingAllWhitespaces: Self {
filter(\.isWhitespace.negated)
}
mutating func removeAllWhitespaces() {
removeAll(where: \.isWhitespace)
}
}
extension Bool {
var negated: Bool { !self }
}
let textInput = "Line 1 \n Line 2 \n\r"
let result = textInput.removingAllWhitespaces //"Line1Line2"
var test = "Line 1 \n Line 2 \n\r"
test.removeAllWhitespaces()
print(test) // "Line1Line2"
Note: For older Swift versions syntax check edit history
For the sake of completeness this is the Regular Expression version
let string = "What is the most efficient way to remove all the spaces and \n \r \tin a String in Swift"
let stringWithoutWhitespace = string.replacingOccurrences(of: "\\s", with: "", options: .regularExpression)
// -> "WhatisthemostefficientwaytoremoveallthespacesandinaStringinSwift"
For Swift 4:
let myString = "This \n is a st\tri\rng"
let trimmedString = myString.components(separatedBy: .whitespacesAndNewlines).joined()
If by spaces you mean whitespaces, be aware that more than one whitespace character exists, although they all look the same.
The following solution takes that into account:
Swift 5:
extension String {
func removingAllWhitespaces() -> String {
return removingCharacters(from: .whitespaces)
}
func removingCharacters(from set: CharacterSet) -> String {
var newString = self
newString.removeAll { char -> Bool in
guard let scalar = char.unicodeScalars.first else { return false }
return set.contains(scalar)
}
return newString
}
}
let noNewlines = "Hello\nWorld".removingCharacters(from: .newlines)
print(noNewlines)
let noWhitespaces = "Hello World".removingCharacters(from: .whitespaces)
print(noWhitespaces)
Use this:
let aString: String = "This is my string"
let newString = aString.stringByReplacingOccurrencesOfString(" ", withString: "", options:[], range: nil)
print(newString)
Output :
Thisismystring
If anyone is wondering why, despite of putting "\n" and "\r" into the set, "\r\n" is not removed from the string, it's because "\r\n" is treated by swift as one character.
Swift 4:
let text = "\r\n This \n is a st\tri\rng"
let test = String(text.filter { !"\r\n\n\t\r".contains($0) })
"\n" is not duplicated by accident
Suppose that you have this string : "some words \nanother word\n\r here something \tand something like \rmdjsbclsdcbsdilvb \n\rand finally this :)"
here the how to remove all possible space :
let possibleWhiteSpace:NSArray = [" ","\t", "\n\r", "\n","\r"] //here you add other types of white space
var string:NSString = "some words \nanother word\n\r here something \tand something like \rmdjsbclsdcbsdilvb \n\rand finally this :)"
print(string)// initial string with white space
possibleWhiteSpace.enumerateObjectsUsingBlock { (whiteSpace, idx, stop) -> Void in
string = string.stringByReplacingOccurrencesOfString(whiteSpace as! String, withString: "")
}
print(string)//resulting string
Let me know if this respond to your question :)
Swift 4:
let string = "Test\n with an st\tri\rng"
print(string.components(separatedBy: .whitespacesAndNewlines))
// Result: "Test with an string"
I just use this:
stringValue.replacingOccurrences(of: " ", with: "").replacingOccurrences(of: "\n", with: "")
Swift 4:
let text = "This \n is a st\tri\rng"
let cleanedText = text.filter { !" \n\t\r".characters.contains($0) }
I have a string composed of words, some of which contain punctuation, which I would like to remove, but I have been unable to figure out how to do this.
For example if I have something like
var words = "Hello, this : is .. a string?"
I would like to be able to create an array with
"[Hello, this, is, a, string]"
My original thought was to use something like words.stringByTrimmingCharactersInSet() to remove any characters I didn't want but that would only take characters off the ends.
I thought maybe I could iterate through the string with something in the vein of
for letter in words {
if NSCharacterSet.punctuationCharacterSet.characterIsMember(letter){
//remove that character from the string
}
}
but I'm unsure how to remove the character from the string. I'm sure there are some problems with the way that if statement is set up, as well, but it shows my thought process.
Xcode 11.4 • Swift 5.2 or later
extension StringProtocol {
var words: [SubSequence] {
split(whereSeparator: \.isLetter.negation)
}
}
extension Bool {
var negation: Bool { !self }
}
let sentence = "Hello, this : is .. a string?"
let words = sentence.words // ["Hello", "this", "is", "a", "string"]
String has a enumerateSubstringsInRange() method.
With the .ByWords option, it detects word boundaries and
punctuation automatically:
Swift 3/4:
let string = "Hello, this : is .. a \"string\"!"
var words : [String] = []
string.enumerateSubstrings(in: string.startIndex..<string.endIndex,
options: .byWords) {
(substring, _, _, _) -> () in
words.append(substring!)
}
print(words) // [Hello, this, is, a, string]
Swift 2:
let string = "Hello, this : is .. a \"string\"!"
var words : [String] = []
string.enumerateSubstringsInRange(string.characters.indices,
options: .ByWords) {
(substring, _, _, _) -> () in
words.append(substring!)
}
print(words) // [Hello, this, is, a, string]
This works with Xcode 8.1, Swift 3:
First define general-purpose extension for filtering by CharacterSet:
extension String {
func removingCharacters(inCharacterSet forbiddenCharacters:CharacterSet) -> String
{
var filteredString = self
while true {
if let forbiddenCharRange = filteredString.rangeOfCharacter(from: forbiddenCharacters) {
filteredString.removeSubrange(forbiddenCharRange)
}
else {
break
}
}
return filteredString
}
}
Then filter using punctuation:
let s:String = "Hello, world!"
s.removingCharacters(inCharacterSet: CharacterSet.punctuationCharacters) // => "Hello world"
let charactersToRemove = NSCharacterSet.punctuationCharacterSet().invertedSet
let aWord = "".join(words.componentsSeparatedByCharactersInSet(charactersToRemove))
An alternate way to filter characters from a set and obtain an array of words is by using the array's filter and reduce methods. It's not as compact as other answers, but it shows how the same result can be obtained in a different way.
First define an array of the characters to remove:
let charactersToRemove = Set(Array(".:?,"))
next convert the input string into an array of characters:
let arrayOfChars = Array(words)
Now we can use reduce to build a string, obtained by appending the elements from arrayOfChars, but skipping all the ones included in charactersToRemove:
let filteredString = arrayOfChars.reduce("") {
let str = String($1)
return $0 + (charactersToRemove.contains($1) ? "" : str)
}
This produces a string without the punctuation characters (as defined in charactersToRemove).
The last 2 steps:
split the string into an array of words, using the blank character as separator:
let arrayOfWords = filteredString.componentsSeparatedByString(" ")
last, remove all empty elements:
let finalArrayOfWords = arrayOfWords.filter { $0.isEmpty == false }
NSScaner way:
let words = "Hello, this : is .. a string?"
//
let scanner = NSScanner(string: words)
var wordArray:[String] = []
var word:NSString? = ""
while(!scanner.atEnd) {
var sr = scanner.scanCharactersFromSet(NSCharacterSet(charactersInString: "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKMNOPQRSTUVWXYZ"), intoString: &word)
if !sr {
scanner.scanLocation++
continue
}
wordArray.append(String(word!))
}
println(wordArray)
I know how to programmatically do it, but I'm sure there's a built-in way...
Every language I've used has some sort of default textual representation for a collection of objects that it will spit out when you try to concatenate the Array with a string, or pass it to a print() function, etc. Does Apple's Swift language have a built-in way of easily turning an Array into a String, or do we always have to be explicit when stringifying an array?
If the array contains strings, you can use the String's join method:
var array = ["1", "2", "3"]
let stringRepresentation = "-".join(array) // "1-2-3"
In Swift 2:
var array = ["1", "2", "3"]
let stringRepresentation = array.joinWithSeparator("-") // "1-2-3"
This can be useful if you want to use a specific separator (hypen, blank, comma, etc).
Otherwise you can simply use the description property, which returns a string representation of the array:
let stringRepresentation = [1, 2, 3].description // "[1, 2, 3]"
Hint: any object implementing the Printable protocol has a description property. If you adopt that protocol in your own classes/structs, you make them print friendly as well
In Swift 3
join becomes joined, example [nil, "1", "2"].flatMap({$0}).joined()
joinWithSeparator becomes joined(separator:) (only available to Array of Strings)
In Swift 4
var array = ["1", "2", "3"]
array.joined(separator:"-")
With Swift 5, according to your needs, you may choose one of the following Playground sample codes in order to solve your problem.
Turning an array of Characters into a String with no separator:
let characterArray: [Character] = ["J", "o", "h", "n"]
let string = String(characterArray)
print(string)
// prints "John"
Turning an array of Strings into a String with no separator:
let stringArray = ["Bob", "Dan", "Bryan"]
let string = stringArray.joined(separator: "")
print(string) // prints: "BobDanBryan"
Turning an array of Strings into a String with a separator between words:
let stringArray = ["Bob", "Dan", "Bryan"]
let string = stringArray.joined(separator: " ")
print(string) // prints: "Bob Dan Bryan"
Turning an array of Strings into a String with a separator between characters:
let stringArray = ["car", "bike", "boat"]
let characterArray = stringArray.flatMap { $0 }
let stringArray2 = characterArray.map { String($0) }
let string = stringArray2.joined(separator: ", ")
print(string) // prints: "c, a, r, b, i, k, e, b, o, a, t"
Turning an array of Floats into a String with a separator between numbers:
let floatArray = [12, 14.6, 35]
let stringArray = floatArray.map { String($0) }
let string = stringArray.joined(separator: "-")
print(string)
// prints "12.0-14.6-35.0"
Swift 2.0 Xcode 7.0 beta 6 onwards uses joinWithSeparator() instead of join():
var array = ["1", "2", "3"]
let stringRepresentation = array.joinWithSeparator("-") // "1-2-3"
joinWithSeparator is defined as an extension on SequenceType
extension SequenceType where Generator.Element == String {
/// Interpose the `separator` between elements of `self`, then concatenate
/// the result. For example:
///
/// ["foo", "bar", "baz"].joinWithSeparator("-|-") // "foo-|-bar-|-baz"
#warn_unused_result
public func joinWithSeparator(separator: String) -> String
}
Swift 3
["I Love","Swift"].joined(separator:" ") // previously joinWithSeparator(" ")
Since no one has mentioned reduce, here it is:
[0, 1, 1, 0].map {"\($0)"}.reduce("") { $0 + $1 } // "0110"
In the spirit of functional programming 🤖
In Swift 4
let array:[String] = ["Apple", "Pear ","Orange"]
array.joined(separator: " ")
To change an array of Optional/Non-Optional Strings
//Array of optional Strings
let array : [String?] = ["1",nil,"2","3","4"]
//Separator String
let separator = ","
//flatMap skips the nil values and then joined combines the non nil elements with the separator
let joinedString = array.flatMap{ $0 }.joined(separator: separator)
//Use Compact map in case of **Swift 4**
let joinedString = array.compactMap{ $0 }.joined(separator: separator
print(joinedString)
Here flatMap, compactMap skips the nil values in the array and appends the other values to give a joined string.
Nowadays, in iOS 13+ and macOS 10.15+, we might use ListFormatter:
let formatter = ListFormatter()
let names = ["Moe", "Larry", "Curly"]
if let string = formatter.string(from: names) {
print(string)
}
That will produce a nice, natural language string representation of the list. A US user will see:
Moe, Larry, and Curly
It will support any languages for which (a) your app has been localized; and (b) the user’s device is configured. For example, a German user with an app supporting German localization, would see:
Moe, Larry und Curly
Mine works on NSMutableArray with componentsJoinedByString
var array = ["1", "2", "3"]
let stringRepresentation = array.componentsJoinedByString("-") // "1-2-3"
In Swift 2.2 you may have to cast your array to NSArray to use componentsJoinedByString(",")
let stringWithCommas = (yourArray as NSArray).componentsJoinedByString(",")
let arrayTemp :[String] = ["Mani","Singh","iOS Developer"]
let stringAfterCombining = arrayTemp.componentsJoinedByString(" ")
print("Result will be >>> \(stringAfterCombining)")
Result will be >>> Mani Singh iOS Developer
If you want to ditch empty strings in the array.
["Jet", "Fire"].filter { !$0.isEmpty }.joined(separator: "-")
If you want to filter nil values as well:
["Jet", nil, "", "Fire"].flatMap { $0 }.filter { !$0.isEmpty }.joined(separator: "-")
you can use joined() to get a single String when you have array of struct also.
struct Person{
let name:String
let contact:String
}
You can easily produce string using map() & joined()
PersonList.map({"\($0.name) - \($0.contact)"}).joined(separator: " | ")
output:
Jhon - 123 | Mark - 456 | Ben - 789
if you want convert custom object array to string or comma separated string (csv) you can use
var stringIds = (self.mylist.map{$0.id ?? 0}).map{String($0)}.joined(separator: ",")
credit to : urvish modi
post: Convert an array of Ints to a comma separated string
A separator can be a bad idea for some languages like Hebrew or Japanese.
Try this:
// Array of Strings
let array: [String] = ["red", "green", "blue"]
let arrayAsString: String = array.description
let stringAsData = arrayAsString.data(using: String.Encoding.utf16)
let arrayBack: [String] = try! JSONDecoder().decode([String].self, from: stringAsData!)
For other data types respectively:
// Set of Doubles
let set: Set<Double> = [1, 2.0, 3]
let setAsString: String = set.description
let setStringAsData = setAsString.data(using: String.Encoding.utf16)
let setBack: Set<Double> = try! JSONDecoder().decode(Set<Double>.self, from: setStringAsData!)
You can either use loops for getting this done. Or by using map.
By mapping:
let array = ["one" , "two" , "three"]
array.map({$0}).joined(seperator : ",")
so in separator you can modify the string.
Output-> ("one,two,three")
The Swift equivalent to what you're describing is string interpolation. If you're thinking about things like JavaScript doing "x" + array, the equivalent in Swift is "x\(array)".
As a general note, there is an important difference between string interpolation vs the Printable protocol. Only certain classes conform to Printable. Every class can be string interpolated somehow. That's helpful when writing generic functions. You don't have to limit yourself to Printable classes.
You can print any object using the print function
or use \(name) to convert any object to a string.
Example:
let array = [1,2,3,4]
print(array) // prints "[1,2,3,4]"
let string = "\(array)" // string == "[1,2,3,4]"
print(string) // prints "[1,2,3,4]"
Create extension for an Array:
extension Array {
var string: String? {
do {
let data = try JSONSerialization.data(withJSONObject: self, options: [.prettyPrinted])
return String(data: data, encoding: .utf8)
} catch {
return nil
}
}
}
if you have string array list , then convert to Int
let arrayList = list.map { Int($0)!}
arrayList.description
it will give you string value
for any Element type
extension Array {
func joined(glue:()->Element)->[Element]{
var result:[Element] = [];
result.reserveCapacity(count * 2);
let last = count - 1;
for (ix,item) in enumerated() {
result.append(item);
guard ix < last else{ continue }
result.append(glue());
}
return result;
}
}
Try This:
let categories = dictData?.value(forKeyPath: "listing_subcategories_id") as! NSMutableArray
let tempArray = NSMutableArray()
for dc in categories
{
let dictD = dc as? NSMutableDictionary
tempArray.add(dictD?.object(forKey: "subcategories_name") as! String)
}
let joinedString = tempArray.componentsJoined(by: ",")
use this when you want to convert list of struct type into string
struct MyStruct {
var name : String
var content : String
}
let myStructList = [MyStruct(name: "name1" , content: "content1") , MyStruct(name: "name2" , content: "content2")]
and covert your array like this way
let myString = myStructList.map({$0.name}).joined(separator: ",")
will produce ===> "name1,name2"
FOR SWIFT 3:
func textField(textField: UITextField, shouldChangeCharactersInRange range: NSRange, replacementString string: String) -> Bool {
if textField == phoneField
{
let newString = NSString(string: textField.text!).replacingCharacters(in: range, with: string)
let components = newString.components(separatedBy: NSCharacterSet.decimalDigits.inverted)
let decimalString = NSString(string: components.joined(separator: ""))
let length = decimalString.length
let hasLeadingOne = length > 0 && decimalString.character(at: 0) == (1 as unichar)
if length == 0 || (length > 10 && !hasLeadingOne) || length > 11
{
let newLength = NSString(string: textField.text!).length + (string as NSString).length - range.length as Int
return (newLength > 10) ? false : true
}
var index = 0 as Int
let formattedString = NSMutableString()
if hasLeadingOne
{
formattedString.append("1 ")
index += 1
}
if (length - index) > 3
{
let areaCode = decimalString.substring(with: NSMakeRange(index, 3))
formattedString.appendFormat("(%#)", areaCode)
index += 3
}
if length - index > 3
{
let prefix = decimalString.substring(with: NSMakeRange(index, 3))
formattedString.appendFormat("%#-", prefix)
index += 3
}
let remainder = decimalString.substring(from: index)
formattedString.append(remainder)
textField.text = formattedString as String
return false
}
else
{
return true
}
}
If you question is something like this:
tobeFormattedString = ["a", "b", "c"]
Output = "abc"
String(tobeFormattedString)