Check if a string exists in an array case insensitively

Question

Declaration:

let listArray = ["kashif"] let word = "kashif" 

then this

contains(listArray, word)  

Returns true but if declaration is:

let word = "Kashif" 

then it returns false because comparison is case sensitive.

How to make this comparison case insensitive?

Answer 1

Xcode 8 • Swift 3 or later

let list = ["kashif"] let word = "Kashif"  if list.contains(where: {$0.caseInsensitiveCompare(word) == .orderedSame}) {     print(true)  // true } 

alternatively:

if list.contains(where: {$0.compare(word, options: .caseInsensitive) == .orderedSame}) {     print(true)  // true } 

if you would like to know the position(s) of the element in the array (it might find more than one element that matches the predicate):

let indices = list.indices.filter { list[$0].caseInsensitiveCompare(word) == .orderedSame } print(indices)  // [0] 

---

You can also use localizedStandardContains method which is case and diacritic insensitive and would match a substring as well:

func localizedStandardContains<T>(_ string: T) -> Bool where T : StringProtocol 

Discussion This is the most appropriate method for doing user-level string searches, similar to how searches are done generally in the system. The search is locale-aware, case and diacritic insensitive. The exact list of search options applied may change over time.

let list = ["kashif"] let word = "Káshif"  if list.contains(where: {$0.localizedStandardContains(word) }) {     print(true)  // true }