Cannot convert value of type 'subSequence' (aka 'String.CharacterView') to type 'String' in collection

Question

I build a webservis that requesting image url from the database. And I want to show it on the swift. But I'm gettig this error on the var photo line:

Cannot convert value of type 'subSequence' (aka 'String.CharacterView') to type 'String' in collection

let requestResponse = self.sendToServer(postUrl: "localhost",data:"abc")

let Seperated = requestResponse.characters.split(separator: " ")

var photo = Seperated[0] as String

let imageURL = URL(string: photo)
if let imageData = try? Data(contentsOf: imageURL!) {
    ...
}

split method on the characters returns aa array of String CharacterView SubSequences. You need to initialize a new string with one of them — Leo Dabus, May 19 '17 at 20:28

score 50 · Answer 1 · answered May 19 '17 at 20:36

50

Consider something like this:

let requestResponse = "some string"
let separated = requestResponse.characters.split(separator: " ")

if let some = separated.first {
    let value = String(some)

    // Output: "some"
}

answered May 19 '17 at 20:36

Michael

697
5
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1

How to get "string" from requestResponse? – Satyam Dec 06 '17 at 17:24
@Satyam see https://stackoverflow.com/questions/32305891/index-of-a-substring-in-a-string-with-swift. You'll just need to adjust a bit – Micah Montoya Dec 07 '17 at 14:38

score 42 · Answer 2 · edited Mar 08 '21 at 00:54

42

The easiest solution.

let separated = requestResponse.characters.split(separator: " ").map { String($0) }

edited Mar 08 '21 at 00:54

Daniel Storm

18,301
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answered May 31 '19 at 13:37

Mickey Mouse

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4

It took me a while to understand why this was neccesary, but it's because split returns a 'substring' object not a string. It's similar but for speed reasons if you modify a substring it modifies from the original string so it's more memory efficient. So what map does is for each 'substring' converts it to an actual string, which is necessary for a lot of functions. Hope this helps someone else understand why this works some more. – Joseph Astrahan Feb 28 '20 at 02:22

score 9 · Answer 3 · answered Oct 16 '18 at 14:22

9

If swift 4 you use like this:

let separated = requestResponse.split(separator: " ")
var photo = String(separated.first ?? "")

answered Oct 16 '18 at 14:22

Guillodacosta

383
3
8

score 9 · Answer 4 · answered May 20 '20 at 14:06

9

split returns a list of Substrings, so we have to map (convert) them to Strings. To do that you can just pass the Substring to the Strings init function:

let separated = requestResponse.characters.split(separator: " ").map(String.init)

answered May 20 '20 at 14:06

Joony

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swift is not a _bad_ language but it does make many simple things frustrating – WestCoastProjects Jun 18 '20 at 13:25
@javadba Gotta agree, mate! :D – N a y y a r Sep 15 '20 at 21:30
1

Swift's tight typing is wonderful, because it prevents so many mistakes that Objective-C's complete looseness allowed. However, it does get very frustrating at times (just let me force you to do this - aargh!). – Carl Smith Jun 27 '21 at 05:52

score 5 · Answer 5 · edited Nov 23 '19 at 01:32

5

An approach to obtain strings from splitting a string using high order functions:

let splitStringArray = requestResponse.split(separator: " ").map({ (substring) in
    return String(substring)
})

edited Nov 23 '19 at 01:32

pkamb

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answered Jan 28 '19 at 20:01

Juan Ramirez

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4

Cannot convert value of type 'subSequence' (aka 'String.CharacterView') to type 'String' in collection

5 Answers5