Reputation: 21902
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?
Upvotes: 450
Views: 389248
Reputation: 72790
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 (hyphen, 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 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:"-")
Upvotes: 835
Reputation: 1423
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")
Upvotes: 6
Reputation: 99
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
Upvotes: 4
Reputation: 2698
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
Upvotes: 5
Reputation: 438307
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
Upvotes: 9
Reputation: 1009
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"
Upvotes: 0
Reputation: 5619
[0, 1, 1, 0].map {"\($0)"}.reduce("") { $0 + $1 } // "0110"
In the spirit of functional programming 🤖
Upvotes: 21
Reputation: 119
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: ",")
Upvotes: 0
Reputation: 560
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;
}
}
Upvotes: 0
Reputation: 92599
With Swift 5, according to your needs, you may choose one of the following Playground sample codes in order to solve your problem.
Character
s into a String
with no separator:let characterArray: [Character] = ["J", "o", "h", "n"]
let string = String(characterArray)
print(string)
// prints "John"
String
s into a String
with no separator:let stringArray = ["Bob", "Dan", "Bryan"]
let string = stringArray.joined(separator: "")
print(string) // prints: "BobDanBryan"
String
s into a String
with a separator between words:let stringArray = ["Bob", "Dan", "Bryan"]
let string = stringArray.joined(separator: " ")
print(string) // prints: "Bob Dan Bryan"
String
s 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"
Float
s 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"
Upvotes: 178
Reputation: 451
if you have string array list , then convert to Int
let arrayList = list.map { Int($0)!}
arrayList.description
it will give you string value
Upvotes: 1
Reputation: 1117
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!)
Upvotes: 2
Reputation: 2923
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.
Upvotes: 9
Reputation: 329
If you question is something like this: tobeFormattedString = ["a", "b", "c"] Output = "abc"
String(tobeFormattedString)
Upvotes: -1
Reputation: 843
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: "-")
Upvotes: 3
Reputation: 2727
In Swift 4
let array:[String] = ["Apple", "Pear ","Orange"]
array.joined(separator: " ")
Upvotes: 16
Reputation: 61880
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
}
}
}
Upvotes: 1
Reputation: 99
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
}
}
Upvotes: -1
Reputation: 1789
Swift 3
["I Love","Swift"].joined(separator:" ") // previously joinWithSeparator(" ")
Upvotes: 23
Reputation: 407
In Swift 2.2 you may have to cast your array to NSArray to use componentsJoinedByString(",")
let stringWithCommas = (yourArray as NSArray).componentsJoinedByString(",")
Upvotes: 4
Reputation: 1456
let arrayTemp :[String] = ["Mani","Singh","iOS Developer"]
let stringAfterCombining = arrayTemp.componentsJoinedByString(" ")
print("Result will be >>> \(stringAfterCombining)")
Result will be >>> Mani Singh iOS Developer
Upvotes: 3
Reputation: 862
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]"
Upvotes: 1
Reputation: 1808
Mine works on NSMutableArray with componentsJoinedByString
var array = ["1", "2", "3"]
let stringRepresentation = array.componentsJoinedByString("-") // "1-2-3"
Upvotes: 4
Reputation: 391
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
}
Upvotes: 49
Reputation: 299633
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.
Upvotes: 1