Wie konvertiere ich eine JSON-Zeichenfolge in ein Wörterbuch?

177

Ich möchte eine Funktion in meinem schnellen Projekt erstellen, die String in das Dictionary-JSON-Format konvertiert, aber ich habe einen Fehler erhalten:

Der Ausdruckstyp kann nicht konvertiert werden (@lvalue NSData, Optionen: IntegerLitralConvertible ...

Das ist mein Code:

func convertStringToDictionary (text:String) -> Dictionary<String,String> {

    var data :NSData = text.dataUsingEncoding(NSUTF8StringEncoding)!
    var json :Dictionary = NSJSONSerialization.JSONObjectWithData(data, options:0, error: nil)
    return json
} 

Ich mache diese Funktion in Objective-C:

- (NSDictionary*)convertStringToDictionary:(NSString*)string {
  NSError* error;
  //giving error as it takes dic, array,etc only. not custom object.
  NSData *data = [string dataUsingEncoding:NSUTF8StringEncoding];
  id json = [NSJSONSerialization JSONObjectWithData:data options:0 error:&error];
  return json;
}
Hose
quelle
Was ist der Wert und Ket, den Sie im Wörterbuch wollen?
Amit89
1
Ich aktualisiere meine Frage ... @ Amit89 egal was Wert und Schlüssel ist !!!
Trousout
@ Amit89 Ich möchte, wenn Text zur Funktion geben json Wörterbuch Format zurückgeben
Trousout
let jsonDictionary = try JSONSerialization.jsonObject(with: jsonData, options: []) as! [String: String];
Vahid Amiri

Antworten:

382

Warnung: Dies ist eine praktische Methode zum Konvertieren einer JSON-Zeichenfolge in ein Wörterbuch, wenn Sie aus irgendeinem Grund von einer JSON-Zeichenfolge aus arbeiten müssen. Wenn Sie jedoch über die JSON- Daten verfügen , sollten Sie stattdessen mit den Daten arbeiten , ohne überhaupt eine Zeichenfolge zu verwenden.

Swift 3

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

let str = "{\"name\":\"James\"}"

let dict = convertToDictionary(text: str)

Swift 2

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

let str = "{\"name\":\"James\"}"

let result = convertStringToDictionary(str)

Original Swift 1 Antwort:

func convertStringToDictionary(text: String) -> [String:String]? {
    if let data = text.dataUsingEncoding(NSUTF8StringEncoding) {
        var error: NSError?
        let json = NSJSONSerialization.JSONObjectWithData(data, options: NSJSONReadingOptions.allZeros, error: &error) as? [String:String]
        if error != nil {
            println(error)
        }
        return json
    }
    return nil
}

let str = "{\"name\":\"James\"}"

let result = convertStringToDictionary(str) // ["name": "James"]

if let name = result?["name"] { // The `?` is here because our `convertStringToDictionary` function returns an Optional
    println(name) // "James"
}

In Ihrer Version haben Sie nicht die richtigen Parameter übergeben NSJSONSerializationund vergessen, das Ergebnis zu übertragen. Es ist auch besser, nach möglichen Fehlern zu suchen. Letzte Anmerkung: Dies funktioniert nur, wenn Ihr Wert ein String ist. Wenn es sich um einen anderen Typ handeln könnte, ist es besser, die Wörterbuchkonvertierung wie folgt zu deklarieren:

let json = NSJSONSerialization.JSONObjectWithData(data, options: NSJSONReadingOptions.allZeros, error: &error) as? [String:AnyObject]

und natürlich müssten Sie auch den Rückgabetyp der Funktion ändern:

func convertStringToDictionary(text: String) -> [String:AnyObject]? { ... }
Eric Aya
quelle
52

Ich habe Eric Ds Antwort für Swift 2 aktualisiert :

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
}
Chris Byatt
quelle
Dieses Beispiel gibt nil zurück. let x = "{" a ": [" b ":" B "," c ":" C "]}" let result = convertStringToDictionary (x)
Syed Tariq
23

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")
    }
}
Danut Pralea
quelle
15

Mit Swift 3 JSONSerializationhat eine Methode aufgerufen json​Object(with:​options:​). json​Object(with:​options:​)hat die folgende Erklärung:

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

Gibt ein Foundation-Objekt aus angegebenen JSON-Daten zurück.

Wenn Sie json​Object(with:​options:​), müssen Sie mit der Fehlerbehandlung beschäftigen ( try, try?oder try!) und Typ Gießen (aus Any). Daher können Sie Ihr Problem mit einem der folgenden Muster lösen.


# 1. Verwenden einer Methode, die einen nicht optionalen Typ auslöst und zurückgibt

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] ?? [:]
}

Verwendung:

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. Verwenden einer Methode, die einen optionalen Typ auslöst und zurückgibt

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]
}

Verwendung:

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. Verwenden einer Methode, die keinen nicht optionalen Typ auslöst und zurückgibt

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] ?? [:]
}

Verwendung:

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. Verwenden einer Methode, die keinen optionalen Typ auslöst und zurückgibt

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]
}

Verwendung:

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
Imanou Petit
quelle
7

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
    }
}
WM
quelle
2
Nur eine weitere Kopie einfügen
J. Doe
1
Wenn Sie keine Optionen verwenden, schreiben Sie sie einfach nicht: JSONSerialization.jsonObject (mit: Daten)
Zaporozhchenko Oleksandr
5

Swift 5

extension String {
    func convertToDictionary() -> [String: Any]? {
        if let data = data(using: .utf8) {
            return try? JSONSerialization.jsonObject(with: data, options: []) as? [String: Any]
        }
        return nil
    }
}
blancoq
quelle
3

Ich habe Code gefunden, der den JSON-String in NSDictionary oder NSArray konvertiert. Fügen Sie einfach die Erweiterung hinzu.

SWIFT 3.0

WIE BENUTZT MAN

let jsonData = (convertedJsonString as! String).parseJSONString

ERWEITERUNG

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
    }
}

}}

Amol Pokale
quelle
Verwenden Sie NSMutable...in Swift überhaupt keine Sammlungstypen. Wenn Sie cast to NSMutable...eingeben , wird niemals ein veränderbares Objekt erstellt. Und ein nicht spezifizierter Typ in Swift 3+ ist Anynicht AnyObject.
Vadian
3

Einzelheiten

  • Xcode Version 10.3 (10G8), Swift 5

Lösung

import Foundation

// MARK: - CastingError

struct CastingError: Error {
    let fromType: Any.Type
    let toType: Any.Type
    init<FromType, ToType>(fromType: FromType.Type, toType: ToType.Type) {
        self.fromType = fromType
        self.toType = toType
    }
}

extension CastingError: LocalizedError {
    var localizedDescription: String { return "Can not cast from \(fromType) to \(toType)" }
}

extension CastingError: CustomStringConvertible { var description: String { return localizedDescription } }

// MARK: - Data cast extensions

extension Data {
    func toDictionary(options: JSONSerialization.ReadingOptions = []) throws -> [String: Any] {
        return try to(type: [String: Any].self, options: options)
    }

    func to<T>(type: T.Type, options: JSONSerialization.ReadingOptions = []) throws -> T {
        guard let result = try JSONSerialization.jsonObject(with: self, options: options) as? T else {
            throw CastingError(fromType: type, toType: T.self)
        }
        return result
    }
}

// MARK: - String cast extensions

extension String {
    func asJSON<T>(to type: T.Type, using encoding: String.Encoding = .utf8) throws -> T {
        guard let data = data(using: encoding) else { throw CastingError(fromType: type, toType: T.self) }
        return try data.to(type: T.self)
    }

    func asJSONToDictionary(using encoding: String.Encoding = .utf8) throws -> [String: Any] {
        return try asJSON(to: [String: Any].self, using: encoding)
    }
}

// MARK: - Dictionary cast extensions

extension Dictionary {
    func toData(options: JSONSerialization.WritingOptions = []) throws -> Data {
        return try JSONSerialization.data(withJSONObject: self, options: options)
    }
}

Verwendung

let value1 = try? data.toDictionary()
let value2 = try? data.to(type: [String: Any].self)
let value3 = try? data.to(type: [String: String].self)
let value4 = try? string.asJSONToDictionary()
let value5 = try? string.asJSON(to: [String: String].self)

Testprobe

Vergessen Sie nicht, den Lösungscode hier einzufügen

func testDescriber(text: String, value: Any) {
    print("\n//////////////////////////////////////////")
    print("-- \(text)\n\n  type: \(type(of: value))\n  value: \(value)")
}

let json1: [String: Any] = ["key1" : 1, "key2": true, "key3" : ["a": 1, "b": 2], "key4": [1,2,3]]
var jsonData = try? json1.toData()
testDescriber(text: "Sample test of func toDictionary()", value: json1)
if let data = jsonData {
    print("  Result: \(String(describing: try? data.toDictionary()))")
}

testDescriber(text: "Sample test of func to<T>() -> [String: Any]", value: json1)
if let data = jsonData {
    print("  Result: \(String(describing: try? data.to(type: [String: Any].self)))")
}

testDescriber(text: "Sample test of func to<T>() -> [String] with cast error", value: json1)
if let data = jsonData {
    do {
        print("  Result: \(String(describing: try data.to(type: [String].self)))")
    } catch {
        print("  ERROR: \(error)")
    }
}

let array = [1,4,5,6]
testDescriber(text: "Sample test of func to<T>() -> [Int]", value: array)
if let data = try? JSONSerialization.data(withJSONObject: array) {
    print("  Result: \(String(describing: try? data.to(type: [Int].self)))")
}

let json2 = ["key1": "a", "key2": "b"]
testDescriber(text: "Sample test of func to<T>() -> [String: String]", value: json2)
if let data = try? JSONSerialization.data(withJSONObject: json2) {
    print("  Result: \(String(describing: try? data.to(type: [String: String].self)))")
}

let jsonString = "{\"key1\": \"a\", \"key2\": \"b\"}"
testDescriber(text: "Sample test of func to<T>() -> [String: String]", value: jsonString)
print("  Result: \(String(describing: try? jsonString.asJSON(to: [String: String].self)))")

testDescriber(text: "Sample test of func to<T>() -> [String: String]", value: jsonString)
print("  Result: \(String(describing: try? jsonString.asJSONToDictionary()))")

let wrongJsonString = "{\"key1\": \"a\", \"key2\":}"
testDescriber(text: "Sample test of func to<T>() -> [String: String] with JSONSerialization error", value: jsonString)
do {
    let json = try wrongJsonString.asJSON(to: [String: String].self)
    print("  Result: \(String(describing: json))")
} catch {
    print("  ERROR: \(error)")
}

Testprotokoll

//////////////////////////////////////////
-- Sample test of func toDictionary()

  type: Dictionary<String, Any>
  value: ["key4": [1, 2, 3], "key2": true, "key3": ["a": 1, "b": 2], "key1": 1]
  Result: Optional(["key4": <__NSArrayI 0x600002a35380>(
1,
2,
3
)
, "key2": 1, "key3": {
    a = 1;
    b = 2;
}, "key1": 1])

//////////////////////////////////////////
-- Sample test of func to<T>() -> [String: Any]

  type: Dictionary<String, Any>
  value: ["key4": [1, 2, 3], "key2": true, "key3": ["a": 1, "b": 2], "key1": 1]
  Result: Optional(["key4": <__NSArrayI 0x600002a254d0>(
1,
2,
3
)
, "key2": 1, "key1": 1, "key3": {
    a = 1;
    b = 2;
}])

//////////////////////////////////////////
-- Sample test of func to<T>() -> [String] with cast error

  type: Dictionary<String, Any>
  value: ["key4": [1, 2, 3], "key2": true, "key3": ["a": 1, "b": 2], "key1": 1]
  ERROR: Can not cast from Array<String> to Array<String>

//////////////////////////////////////////
-- Sample test of func to<T>() -> [Int]

  type: Array<Int>
  value: [1, 4, 5, 6]
  Result: Optional([1, 4, 5, 6])

//////////////////////////////////////////
-- Sample test of func to<T>() -> [String: String]

  type: Dictionary<String, String>
  value: ["key1": "a", "key2": "b"]
  Result: Optional(["key1": "a", "key2": "b"])

//////////////////////////////////////////
-- Sample test of func to<T>() -> [String: String]

  type: String
  value: {"key1": "a", "key2": "b"}
  Result: Optional(["key1": "a", "key2": "b"])

//////////////////////////////////////////
-- Sample test of func to<T>() -> [String: String]

  type: String
  value: {"key1": "a", "key2": "b"}
  Result: Optional(["key1": a, "key2": b])

//////////////////////////////////////////
-- Sample test of func to<T>() -> [String: String] with JSONSerialization error

  type: String
  value: {"key1": "a", "key2": "b"}
  ERROR: Error Domain=NSCocoaErrorDomain Code=3840 "Invalid value around character 21." UserInfo={NSDebugDescription=Invalid value around character 21.}
Wassili Bodnarchuk
quelle
0

Für Swift 5 schreibe ich eine Demo, um sie zu überprüfen.

extension String {

    /// convert JsonString to Dictionary
    func convertJsonStringToDictionary() -> [String: Any]? {
        if let data = data(using: .utf8) {
            return (try? JSONSerialization.jsonObject(with: data, options: [])) as? [String: Any]
        }

        return nil
    }
}

let str = "{\"name\":\"zgpeace\"}"
let dict = str.convertJsonStringToDictionary()
print("string > \(str)")  
// string > {"name":"zgpeace"}
print("dicionary > \(String(describing: dict))") 
// dicionary > Optional(["name": zgpeace])
Zgpeace
quelle
-1
let JSONData = jsonString.data(using: .utf8)!

let jsonResult = try JSONSerialization.jsonObject(with: data, options: .mutableLeaves)

guard let userDictionary = jsonResult as? Dictionary<String, AnyObject> else {
            throw NSError()}
t.ios
quelle