let string = "hello hi" var hello = "" var hi = "" I wan't to split my string so that the value of hello get "hello" and the value of hi get "hi"
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2 Answers
Try this:
var myString: String = "hello hi"; var myStringArr = myString.componentsSeparatedByString(" ") Where myString is the name of your string, and myStringArr contains the components separated by the space.
Then you can get the components as:
var hello: String = myStringArr [0] var hi: String = myStringArr [1] Doc: componentsSeparatedByString
EDIT: For Swift 3, the above will be:
var myStringArr = myString.components(separatedBy: " ") 
1 Comment
Dave
The above has changed in Swift 3, it is now: 'var arrayOfSeperatedParts = myOriginalString.components(separatedBy: "insertSeperatorHere")'
Here are split that receives regex as well. You can define extension for future usage:
Swift 4
extension String { func split(regex pattern: String) -> [String] { guard let re = try? NSRegularExpression(pattern: pattern, options: []) else { return [] } let nsString = self as NSString // needed for range compatibility let stop = "<SomeStringThatYouDoNotExpectToOccurInSelf>" let modifiedString = re.stringByReplacingMatches( in: self, options: [], range: NSRange(location: 0, length: nsString.length), withTemplate: stop) return modifiedString.components(separatedBy: stop) } } Examples:
let string1 = "hello world" string1.split(regex: " ") // ["hello", "world"] let string2 = "hello world" string2.split(regex: "[ ]+") // ["hello", "world"] Swift 2.2
extension String { func split(regex pattern: String) -> [String] { guard let re = try? NSRegularExpression(pattern: pattern, options: []) else { return [] } let nsString = self as NSString // needed for range compatibility let stop = "<SomeStringThatYouDoNotExpectToOccurInSelf>" let modifiedString = re.stringByReplacingMatchesInString( self, options: [], range: NSRange(location: 0, length: nsString.length), withTemplate: stop) return modifiedString.componentsSeparatedByString(stop) } } Swift 2.0
extension String { // java, javascript, PHP use 'split' name, why not in Swift? :) func split(regex: String) -> Array<String> { do{ let regEx = try NSRegularExpression(pattern: regex, options: NSRegularExpressionOptions()) let stop = "<SomeStringThatYouDoNotExpectToOccurInSelf>" let nsString = self as NSString // needed for range compatibility let modifiedString = regEx.stringByReplacingMatchesInString (self, options: NSMatchingOptions(), range: NSRange(location: 0, length: nsString.length), withTemplate:stop) return modifiedString.componentsSeparatedByString(stop) } catch { return [] } } } Swift 1.1
extension String { // java, javascript, PHP use 'split' name, why not in Swift? :) func split(splitter: String) -> Array<String> { let regEx = NSRegularExpression(pattern: splitter, options: NSRegularExpressionOptions(), error: nil) let stop = "<SomeStringThatYouDoNotExpectToOccurInSelf>" let modifiedString = regEx.stringByReplacingMatchesInString (self, options: NSMatchingOptions(), range: NSMakeRange(0, countElements(self)), withTemplate:stop) return modifiedString.componentsSeparatedByString(stop) } } Examples:
let string1 = "hello world" string1.split(" ") // ["hello", "world"] let string2 = "hello world" string2.split("[ ]+") // ["hello", "world"] 
4 Comments
Maxim Shoustin
@downvoter please describe
Tyler Liu
I didn't downvote. I know the disadvantage of the method: you don't known when the parameter is a plain string while when it is a regex pattern string. For example, what if I want to split a string by "\d+" (literal string, not regex)
masukomi
@TylerLong the method tells you it's a regexp. If you want to split by the literal "\d+" then you pass in a regex that describes that literal string. E.g.:
"\\\\d\\+"
ctietze
I'd suggest to rename the method to
func split(regex pattern: String) -> [String] so on the call-site it's string1.split(regex: " ") instead. ("pattern" because the regex is supposed to be NSRegularExpression) The existing CollectionType.split can take a single character, too, so it's hard to tell which one is called when the parameter is " " only.