How to parse array of strings in swift?

Question

I am writing a simple full stack app and I made the return from the backend to be, in certain cases, just like an array of strings like below:

["one","two","three"]

The problem I am having now is that this return can't be parsed in swift like usual JSON data. I searched but found nothing. What I am trying to do is :

let json = try JSONSerialization.jsonObject(with: data!, options: [.mutableContainers, .allowFragments]) as? [String]

Unfortunately, this doesn't work at all How can I parse this just like a normal array?

Thanks in advance.

UPDATE: I am adding some more relevant code below:

let task = URLSession.shared.dataTask(with: request as URLRequest){ data, response, error in { do { let json = try JSONSerialization.jsonObject(with: data!, options: [.mutableContainers, .allowFragments]) as? [String] print(json) }catch{ print(error) } } datatask.resume()

The error I am getting in the console is "Error Domain=NSCocoaErrorDomain Code=3840 "Invalid value around character 0." UserInfo={NSDebugDescription=Invalid value around character 0.}"

Thanks for all of you.

Show more relevant code. Where does data come from? What is the actual value you are trying to parse? — rmaddy
– rmaddy, Commented Jan 20, 2018 at 18:12
@vadian The console then yells at me complaining that the json return is neither an array nor a dictionary. — Badr
– Badr, Commented Jan 20, 2018 at 21:21

vadian · Accepted Answer · 2018-01-20 22:03:12Z

If you're getting the 3840 error, then the JSON is not valid JSON, period.

The JSON string equivalent of the array is supposed to be

let jsonString = "[\"one\",\"two\",\"three\"]"

The Swift 4 literal multiline syntax shows the actual format without the escaping backslashes

let jsonString = """ ["one","two","three"] """

You are able to parse it without any options (no .allowFragments, and no .mutableContainers)

let data = Data(jsonString.utf8) do { let array = try JSONSerialization.jsonObject(with: data) as! [String] print(array) // ["one", "two", "three"] } catch { print(error) }

Almost everybody misuses the JSONSerialization Reading Options

.allowFragments is only needed if the root object is not array and not dictionary
.mutableContainers is completely meaningless in Swift

In 99% of the cases you can omit the options parameter

Thanks for the answer. Do you have any idea what's that equivalent to in php as I did try to test the return using postman and it seemed fine?
The php equivalent is $array = array('one', 'two','three');
This is what I do! I use array_push($array, $row->Name) where $row is $row = mysqli_fetch_object($result)
Please create a string let string = String(data: data!, encoding: .utf8) before the JSONSerialization line and print and post that.

Amrit Trivedi · Accepted Answer · 2018-01-21 19:01:12Z

You have to give proper JSON response as data. After that, JSON serialisation will work properly. Your JSON response should be like this:

{ "array": [ "one", "two", "three" ] } let json = try JSONSerialization.jsonObject(with: data!, options: [.mutableContainers, .allowFragments]) as? [String] print(json)

After that in above line you will your array response like this one - http://jsoneditoronline.org/?id=eafd7ff34b45b2a380ebbe5959607906

Update ####

Your response starts with array and it contains dictionary see below code for your example

 let jsonString = """ [{"Name":"one"},{"Name":"two"},{"Name":"three"}] """ if let jsonData = jsonString.data(using: .utf8) { let json = try JSONSerialization.jsonObject(with: jsonData, options: [.mutableContainers, .allowFragments]) as? [[String:String]] print(json) }

I would do that but it doesn't work also for example I have tried a return of : [{"Name":"one"},{"Name":"two"},{"Name":"three"}] But still couldn't parse it as [Dictionary<String,String>]

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