CODEWITHSUNDEEP
jsonxcodeswiftparsing
SpringN
SpringN

Reputation: 954

JSON Parsing in Swift 3

Has anyone been able to find a way to parse through JSON files in Swift 3? I have been able to get the data to return but I am unsuccessful when it comes to breaking the data down into specific fields. I would post sample code but I've gone through so many different methods unsuccessfully and haven't saved any. The basic format I want to parse through is something like this. Thanks in advance.

{
  "Language": {

    "Field":[
          {
          "Number":"976",
          "Name":"Test"
          },
          {
          "Number":"977",
          "Name":"Test"
          }
       ]
   }
}

Upvotes: 21

Views: 54498

Answers (8)

Rizwan Shaikh
Rizwan Shaikh

Reputation: 271

dict = {
    message = "Login successfully.";
    status = 1;
    "user_details" =     (
                {
            dob = "1900-11-18";
            email = "[email protected]";
            gender = male;
            name = Rizwan;
            nickname = Shaikh;
            "profile_pic" = "1483434421.jpeg";
            "social_id" = "<null>";
            "user_id" = 2;
        }
    );
}

We can parse above json in Swift 3 as

var dict2  = dict as! [String : Any]
print(dict);
let demoStr = dict2["message"] as! String
print(demoStr)
let demoArray = dict2["user_details"] as! [Any]
let demoDict = demoArray[0] as! [String:Any]
print(demoDict["dob"]!)

Upvotes: -1

ismailtsn92
ismailtsn92

Reputation: 21

Use SwiftJson library. I think its very easy way to parse.

let count: Int? = json["Field"].array?.count
if let ct = count {            
    for index in 0...ct-1{
        let number = json ["Field"][index]["number"].string
        let name = json ["Field"][index]["name"].string

....

like this .

Upvotes: 0

user8771317
user8771317

Reputation:

JSON Parsing using Swift 4 in Simple WAY

   let url = URL(string: "http://mobileappdevelop.co/TIPIT/webservice/get_my_groups?user_id=5")
    URLSession.shared.dataTask(with:url!, completionHandler: {(data, response, error) in
        guard let data = data, error == nil else { return }

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

             print(json)

            let posts =  json["Field"] as? [[String: Any]] ?? []
            print(posts)
        } catch let error as NSError {
            print(error)
        }

    }).resume()

}

Upvotes: 0

Jeffery Thomas
Jeffery Thomas

Reputation: 42588

Have you tried JSONSerialization.jsonObject(with:options:)?

var jsonString = "{" +
    "\"Language\": {" +
    "\"Field\":[" +
    "{" +
    "\"Number\":\"976\"," +
    "\"Name\":\"Test\"" +
    "}," +
    "{" +
    "\"Number\":\"977\"," +
    "\"Name\":\"Test\"" +
    "}" +
    "]" +
    "}" +
    "}"

var data = jsonString.data(using: .utf8)!

let json = try? JSONSerialization.jsonObject(with: data)

Swift sometimes produces some very odd syntax.

if let number = json?["Language"]??["Field"]??[0]?["Number"] as? String {
    print(number)
}

Everything in the JSON object hierarchy ends up getting wrapped as an optional (ie. AnyObject?). Array<T> subscript returns a non-optional T. For this JSON, which is wrapped in an optional, array subscript returns Optional<AnyObject>. However, Dictionary<K, V> subscript returns an Optional<V>. For this JSON, subscript returns the very odd looking Optional<Optional<AnyObject>> (ie. AnyObject??).

  • json is an Optional<AnyObject>.
  • json?["Language"] returns an Optional<Optional<AnyObject>>.
  • json?["Language"]??["Field"] returns an Optional<Optional<AnyObject>>.
  • json?["Language"]??["Field"]??[0] returns an Optional<AnyObject>.
  • json?["Language"]??["Field"]??[0]?["Number"] returns an Optional<Optional<AnyObject>>.
  • json?["Language"]??["Field"]??[0]?["Number"] as? String returns an Optional<String>.

The Optional<String> is then used by the if let syntax to product a String.

Final note: iterating the field array looks like this.

for field in json?["Language"]??["Field"] as? [AnyObject] ?? [] {
    if let number = field["Number"] as? String {
        print(number)
    }
}

Swift 4 Update

Swift 4 makes this all much easier to deal with. Again we will start with your test data (""" makes this so much nicer).

let data = """
{
  "Language": {

    "Field":[
          {
          "Number":"976",
          "Name":"Test"
          },
          {
          "Number":"977",
          "Name":"Test"
          }
       ]
   }
}
""".data(using: .utf8)!

Next we can define classes around the objects used in your JSON.

struct Object: Decodable {
    let language: Language
    enum CodingKeys: String, CodingKey { case language="Language" }
}

struct Language: Decodable {
    let fields: [Field]
    enum CodingKeys: String, CodingKey { case fields="Field" }
}

struct Field: Decodable {
    let number: String
    let name: String
    enum CodingKeys: String, CodingKey { case number="Number"; case name="Name" }
}

The CodingKeys enum is how struct properties are mapped to JSON object member strings. This mapping is done automagically by Decodable.

Parsing the JSON now is simple.

let object = try! JSONDecoder().decode(Object.self, from: data)

print(object.language.fields[0].name)

for field in object.language.fields {
    print(field.number)
}

Upvotes: 28

Ashis Laha
Ashis Laha

Reputation: 2364

JSON Parsing in swift 4 using Decodable Protocol :

I create a mocky file using your json object :

http://www.mocky.io/v2/5a280c282f0000f92c0635e6

Here is the code to parse the JSON :

Model Creation :

import UIKit

struct Item : Decodable { 
// Properties must be the same name as specified in JSON , else it will return nil
var Number : String
var Name : String
}

struct Language : Decodable {
 var Field : [Item]
}

struct Result : Decodable {
 var Language : Language
}

You can use optional in the model if you are uncertain that something might be missing in JSON file.

This is the parsing Logic :

class ViewController: UIViewController {

let url = "http://www.mocky.io/v2/5a280c282f0000f92c0635e6"

private func parseJSON() {

    guard let url = URL(string: url) else { return }

    let session = URLSession.shared.dataTask(with: url) { (data, response, error) in
        guard let data = data else { return }
        guard let result = try? JSONDecoder().decode(Result.self, from: data) else { return }
        print("\n\nResult : \(result)")
    }
    session.resume()
}

override func viewDidLoad() {
    super.viewDidLoad()
    parseJSON()
}
}

The Print Output :

 Result : Result(Language: JSON_Parsing.Language(Field: [JSON_Parsing.Item(Number: "976", Name: "Test"), JSON_Parsing.Item(Number: "977", Name: "Test")]))

This the github Project link. You can check.

Upvotes: 0

BhuShan PaWar
BhuShan PaWar

Reputation: 79

 override func viewDidLoad() {
        super.viewDidLoad()
        let url=URL(string:"http://api.androidhive.info/contacts/")
        do {
            let allContactsData = try Data(contentsOf: url!)
            let allContacts = try JSONSerialization.jsonObject(with: allContactsData, options: JSONSerialization.ReadingOptions.allowFragments) as! [String : AnyObject]
            if let arrJSON = allContacts["contacts"] {
                for index in 0...arrJSON.count-1 {
                    let aObject = arrJSON[index] as! [String : AnyObject]
                    names.append(aObject["name"] as! String)
                    contacts.append(aObject["email"] as! String)
                }
            }
            print(names)
            print(contacts)
            self.tableView.reloadData()
        }
        catch {
        }
    }

Upvotes: 0

Gene De Lisa
Gene De Lisa

Reputation: 3838

Shoving JSON into a string manually is a pita. Why don't you just put the JSON into a file and read that in?

Swift 3:

let bundle = Bundle(for: type(of: self))
    if let theURL = bundle.url(forResource: "response", withExtension: "json") {
        do {
            let data = try Data(contentsOf: theURL)
            if let parsedData = try? JSONSerialization.jsonObject(with: data) as! [String:Any] {
                grok(parsedData)
            }
        } catch {
            print(error)
        }
    }

Upvotes: 3

serg_zhd
serg_zhd

Reputation: 1043

In Xcode 8 and Swift 3 id now imports as Any rather than AnyObject

This means that JSONSerialization.jsonObject(with: data) returns Any. So you have to cast the json data to a specific type like [String:Any]. Same applies to the next fields down the json.

var jsonString = "{" +
    "\"Language\": {" +
    "\"Field\":[" +
    "{" +
    "\"Number\":\"976\"," +
    "\"Name\":\"Test1\"" +
    "}," +
    "{" +
    "\"Number\":\"977\"," +
    "\"Name\":\"Test2\"" +
    "}" +
    "]" +
    "}" +
"}"

var data = jsonString.data(using: .utf8)!
if let parsedData = try? JSONSerialization.jsonObject(with: data) as! [String:Any] {
    let language = parsedData["Language"] as! [String:Any]
    print(language)
    let field = language["Field"] as! [[String:Any]]
    let name = field[0]["Name"]!
    print(name) // ==> Test1
}

In practice you would probably want some specific field buried in the json. Lets assume it's the Name field of the first element of Field array. You can use a chain of unwraps like this to safely access the field:

var data = jsonString.data(using: .utf8)!
if let json = try? JSONSerialization.jsonObject(with: data) as? [String:Any],
    let language = json?["Language"] as? [String:Any],
    let field = language["Field"] as? [[String:Any]],
    let name = field[0]["Name"] as? String, field.count > 0 {
    print(name) // ==> Test1
} else {
    print("bad json - do some recovery")
}

Also you may want to check Apple's Swift Blog Working with JSON in Swift

Upvotes: 14

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