Как преобразовать строку JSON в словарь?

я обновил ответ Эрика D для Swift 2:

func convertStringToDictionary(text: String) -> [String:AnyObject]? {
    if let data = text.dataUsingEncoding(NSUTF8StringEncoding) {
        do {
            let json = try NSJSONSerialization.JSONObjectWithData(data, options: .MutableContainers) as? [String:AnyObject]
            return json
        } catch {
            print("Something went wrong")
        }
    }
    return nil
}

Swift 3:

if let data = text.data(using: String.Encoding.utf8) {
    do {
        let json = try JSONSerialization.jsonObject(with: data, options: .mutableContainers) as? [String:Any]
        print(json)
    } catch {
        print("Something went wrong")
    }
}

С Swift 3,JSONSerialization имеет метод под названием json​Object(with:​options:​). json​Object(with:​options:​) есть следующее объявление:

class func jsonObject(with data: Data, options opt: JSONSerialization.ReadingOptions = []) throws -> Any

возвращает объект Foundation из данных JSON.

при использовании json​Object(with:​options:​), вам приходится иметь дело с обработкой ошибок (try, try? или try!) и тип литья (от Any). Таким образом, вы можете решить свою проблему с помощью одного из следующих шаблонов.

#1. Используя метод это бросает и возвращает необязательный тип

import Foundation

func convertToDictionary(from text: String) throws -> [String: String] {
    guard let data = text.data(using: .utf8) else { return [:] }
    let anyResult: Any = try JSONSerialization.jsonObject(with: data, options: [])
    return anyResult as? [String: String] ?? [:]
}

использование:

let string1 = "{\"City\":\"Paris\"}"
do {
    let dictionary = try convertToDictionary(from: string1)
    print(dictionary) // prints: ["City": "Paris"]
} catch {
    print(error)
}

let string2 = "{\"Quantity\":100}"
do {
    let dictionary = try convertToDictionary(from: string2)
    print(dictionary) // prints [:]
} catch {
    print(error)
}

let string3 = "{\"Object\"}"
do {
    let dictionary = try convertToDictionary(from: string3)
    print(dictionary)
} catch {
    print(error) // prints: Error Domain=NSCocoaErrorDomain Code=3840 "No value for key in object around character 9." UserInfo={NSDebugDescription=No value for key in object around character 9.}
}

#2. Использование метода, который выбрасывает и возвращает необязательный тип

import Foundation

func convertToDictionary(from text: String) throws -> [String: String]? {
    guard let data = text.data(using: .utf8) else { return [:] }
    let anyResult: Any = try JSONSerialization.jsonObject(with: data, options: [])
    return anyResult as? [String: String]
}

использование:

let string1 = "{\"City\":\"Paris\"}"
do {
    let dictionary = try convertToDictionary(from: string1)
    print(String(describing: dictionary)) // prints: Optional(["City": "Paris"])
} catch {
    print(error)
}

let string2 = "{\"Quantity\":100}"
do {
    let dictionary = try convertToDictionary(from: string2)
    print(String(describing: dictionary)) // prints nil
} catch {
    print(error)
}

let string3 = "{\"Object\"}"
do {
    let dictionary = try convertToDictionary(from: string3)
    print(String(describing: dictionary))
} catch {
    print(error) // prints: Error Domain=NSCocoaErrorDomain Code=3840 "No value for key in object around character 9." UserInfo={NSDebugDescription=No value for key in object around character 9.}
}

#3. Использование метода, который не бросает и возвращает необязательный тип

import Foundation

func convertToDictionary(from text: String) -> [String: String] {
    guard let data = text.data(using: .utf8) else { return [:] }
    let anyResult: Any? = try? JSONSerialization.jsonObject(with: data, options: [])
    return anyResult as? [String: String] ?? [:]
}

использование:

let string1 = "{\"City\":\"Paris\"}"
let dictionary1 = convertToDictionary(from: string1)
print(dictionary1) // prints: ["City": "Paris"]

let string2 = "{\"Quantity\":100}"
let dictionary2 = convertToDictionary(from: string2)
print(dictionary2) // prints: [:]

let string3 = "{\"Object\"}"
let dictionary3 = convertToDictionary(from: string3)
print(dictionary3) // prints: [:]

#4. Используя метод это не бросает и возвращает необязательный тип

import Foundation

func convertToDictionary(from text: String) -> [String: String]? {
    guard let data = text.data(using: .utf8) else { return nil }
    let anyResult = try? JSONSerialization.jsonObject(with: data, options: [])
    return anyResult as? [String: String]
}

использование:

let string1 = "{\"City\":\"Paris\"}"
let dictionary1 = convertToDictionary(from: string1)
print(String(describing: dictionary1)) // prints: Optional(["City": "Paris"])

let string2 = "{\"Quantity\":100}"
let dictionary2 = convertToDictionary(from: string2)
print(String(describing: dictionary2)) // prints: nil

let string3 = "{\"Object\"}"
let dictionary3 = convertToDictionary(from: string3)
print(String(describing: dictionary3)) // prints: nil

Я нашел код, который преобразует строку json в NSDictionary или NSArray.просто добавьте расширение SWIFT 3.0

КАК ИСПОЛЬЗОВАТЬ

let jsonData = (convertedJsonString as! String).parseJSONString

расширение

 extension String
{
var parseJSONString: AnyObject?
{
    let data = self.data(using: String.Encoding.utf8, allowLossyConversion: false)
    if let jsonData = data
    {
        // Will return an object or nil if JSON decoding fails
        do
        {
            let message = try JSONSerialization.jsonObject(with: jsonData, options:.mutableContainers)
            if let jsonResult = message as? NSMutableArray {
                return jsonResult //Will return the json array output
            } else if let jsonResult = message as? NSMutableDictionary {
                return jsonResult //Will return the json dictionary output
            } else {
                return nil
            }
        }
        catch let error as NSError
        {
            print("An error occurred: \(error)")
            return nil
        }
    }
    else
    {
        // Lossless conversion of the string was not possible
        return nil
    }
}

}

Swift 4

extension String {
    func convertToDictionary() -> [String: Any]? {
        if let data = self.data(using: .utf8) {
            do {
                return try JSONSerialization.jsonObject(with: data, options: []) as? [String: Any]
            } catch {
                print(error.localizedDescription)
            }
        }
        return nil
    }
}