在 Swift 中将 JSON 字符串转换为 Object 的简单而干净的方法

Question

我一直在寻找将相当简单的 JSON 字符串转换为 Swift 中的对象类型的方法，但无济于事。

这是 Web 服务调用的代码：

func GetAllBusiness() {

        Alamofire.request(.GET, "http://MyWebService/").responseString { (request, response, string, error) in

                println(string)

        }
}

我有一个 swift 结构 Business.swift：

struct Business {
    var Id : Int = 0
    var Name = ""
    var Latitude = ""
    var Longitude = ""
    var Address = ""
}

这是我部署的测试服务：

[
  {
    "Id": 1,
    "Name": "A",
    "Latitude": "-35.243256",
    "Longitude": "149.110701",
    "Address": null
  },
  {
    "Id": 2,
    "Name": "B",
    "Latitude": "-35.240592",
    "Longitude": "149.104843",
    "Address": null
  }
  ...
]

如果有人指导我完成这件事，我会很高兴。

谢谢。

Answer 1

以下是一些如何从简单示例开始的提示。

考虑您有以下 JSON 数组字符串（类似于您的），例如：

 var list:Array<Business> = []

  // left only 2 fields for demo
  struct Business {
    var id : Int = 0
    var name = ""               
 }

 var jsonStringAsArray = "[\n" +
        "{\n" +
        "\"id\":72,\n" +
        "\"name\":\"Batata Cremosa\",\n" +            
        "},\n" +
        "{\n" +
        "\"id\":183,\n" +
        "\"name\":\"Caldeirada de Peixes\",\n" +            
        "},\n" +
        "{\n" +
        "\"id\":76,\n" +
        "\"name\":\"Batata com Cebola e Ervas\",\n" +            
        "},\n" +
        "{\n" +
        "\"id\":56,\n" +
        "\"name\":\"Arroz de forma\",\n" +            
    "}]"


        // convert String to NSData
        var data: NSData = jsonStringAsArray.dataUsingEncoding(NSUTF8StringEncoding)!
        var error: NSError?

        // convert NSData to 'AnyObject'
        let anyObj: AnyObject? = NSJSONSerialization.JSONObjectWithData(data, options: NSJSONReadingOptions(0),
            error: &error)
        println("Error: \(error)")

     // convert 'AnyObject' to Array<Business>
     list = self.parseJson(anyObj!)

     //===============

    func parseJson(anyObj:AnyObject) -> Array<Business>{

        var list:Array<Business> = []

         if  anyObj is Array<AnyObject> {

            var b:Business = Business()

            for json in anyObj as Array<AnyObject>{
             b.name = (json["name"] as AnyObject? as? String) ?? "" // to get rid of null
             b.id  =  (json["id"]  as AnyObject? as? Int) ?? 0                 

               list.append(b)
            }// for

        } // if

      return list

    }//func    

[编辑]

去掉 null 改为：

b.name = (json["name"] as AnyObject? as? String) ?? ""
b.id  =  (json["id"]  as AnyObject? as? Int) ?? 0 

另请参阅合并运算符的参考（又名?? ）

希望能帮到你解决问题

Answer 2

快速 3/4

extension String {
    func toJSON() -> Any? {
        guard let data = self.data(using: .utf8, allowLossyConversion: false) else { return nil }
        return try? JSONSerialization.jsonObject(with: data, options: .mutableContainers)
    }
}

示例用法：

 let dict = myString.toJSON() as? [String:AnyObject] // can be any type here

Answer 3

简单的 String 扩展就足够了：

extension String {

    var parseJSONString: AnyObject? {

        let data = self.dataUsingEncoding(NSUTF8StringEncoding, allowLossyConversion: false)

        if let jsonData = data {
            // Will return an object or nil if JSON decoding fails
            return NSJSONSerialization.JSONObjectWithData(jsonData, options: NSJSONReadingOptions.MutableContainers, error: nil)
        } else {
            // Lossless conversion of the string was not possible
            return nil
        }
    }
}

然后：

var jsonString = "[\n" +
    "{\n" +
    "\"id\":72,\n" +
    "\"name\":\"Batata Cremosa\",\n" +            
    "},\n" +
    "{\n" +
    "\"id\":183,\n" +
    "\"name\":\"Caldeirada de Peixes\",\n" +            
    "},\n" +
    "{\n" +
    "\"id\":76,\n" +
    "\"name\":\"Batata com Cebola e Ervas\",\n" +            
    "},\n" +
    "{\n" +
    "\"id\":56,\n" +
    "\"name\":\"Arroz de forma\",\n" +            
"}]"

let json: AnyObject? = jsonString.parseJSONString
println("Parsed JSON: \(json!)")
println("json[3]: \(json![3])")

/* Output:

Parsed JSON: (
    {
    id = 72;
    name = "Batata Cremosa";
    },
    {
    id = 183;
    name = "Caldeirada de Peixes";
    },
    {
    id = 76;
    name = "Batata com Cebola e Ervas";
    },
    {
    id = 56;
    name = "Arroz de forma";
    }
)

json[3]: {
    id = 56;
    name = "Arroz de forma";
}
*/

Answer 4

Swift 4更优雅地解析 JSON。 按照这个简化的示例，只需为您的结构采用可编码的协议：

struct Business: Codable {
    let id: Int
    let name: String
}

要解析 JSON 数组，您需要告诉解码器数据数组的对象是什么

let parsedData = decoder.decode([Business].self, from: data)

这是一个完整的工作示例：

import Foundation

struct Business: Codable {
    let id: Int
    let name: String
}

// Generating the example JSON data: 
let originalObjects = [Business(id: 0, name: "A"), Business(id: 1, name: "B")]
let encoder = JSONEncoder()
let data = try! encoder.encode(originalObjects)

// Parsing the data: 
let decoder = JSONDecoder()
let parsedData = try! decoder.decode([Business].self, from: data)

有关更多背景信息，请查看此优秀指南。

Answer 5

对于Swift 4

我使用了 @ Passkit的逻辑，但我必须按照 Swift 4 进行更新

---

Step.1为字符串类创建扩展

import UIKit


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
                    {
                        print(jsonResult)

                        return jsonResult //Will return the json array 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
            }
        }
    }

---

Step.2这就是我在视图控制器中的使用方式

var jsonString = "[\n" +
    "{\n" +
    "\"id\":72,\n" +
    "\"name\":\"Batata Cremosa\",\n" +            
    "},\n" +
    "{\n" +
    "\"id\":183,\n" +
    "\"name\":\"Caldeirada de Peixes\",\n" +            
    "},\n" +
    "{\n" +
    "\"id\":76,\n" +
    "\"name\":\"Batata com Cebola e Ervas\",\n" +            
    "},\n" +
    "{\n" +
    "\"id\":56,\n" +
    "\"name\":\"Arroz de forma\",\n" +            
"}]"

 //Convert jsonString to jsonArray

let json: AnyObject? = jsonString.parseJSONString
print("Parsed JSON: \(json!)")
print("json[2]: \(json![2])")

所有功劳归于原始用户，我刚刚更新了最新的 swift 版本

Answer 6

我编写了一个库，它使在 Swift 中处理 json 数据和反序列化变得轻而易举。 你可以在这里得到它： https : //github.com/isair/JSONHelper

编辑：我更新了我的图书馆，你现在可以这样做：

class Business: Deserializable {
    var id: Int?
    var name = "N/A"  // This one has a default value.

    required init(data: [String: AnyObject]) {
        id <-- data["id"]
        name <-- data["name"]
    }
}

var businesses: [Business]()

Alamofire.request(.GET, "http://MyWebService/").responseString { (request, response, string, error) in
    businesses <-- string
}

旧答案：

首先，不要使用 .responseString，而是使用 .response 来获取响应对象。 然后将您的代码更改为：

func getAllBusinesses() {

    Alamofire.request(.GET, "http://MyWebService/").response { (request, response, data, error) in
        var businesses: [Business]?

        businesses <-- data

        if businesses == nil {
            // Data was not structured as expected and deserialization failed, do something.
        } else {
            // Do something with your businesses array. 
        }
    }
}

你需要像这样制作一个商业类：

class Business: Deserializable {
    var id: Int?
    var name = "N/A"  // This one has a default value.

    required init(data: [String: AnyObject]) {
        id <-- data["id"]
        name <-- data["name"]
    }
}

您可以在我的 GitHub 存储库上找到完整的文档。 玩得开心！

Answer 7

对于Swift 4 ，我使用Codable协议编写了​​这个扩展：

struct Business: Codable {
    var id: Int
    var name: String
}

extension String {

    func parse<D>(to type: D.Type) -> D? where D: Decodable {

        let data: Data = self.data(using: .utf8)!

        let decoder = JSONDecoder()

        do {
            let _object = try decoder.decode(type, from: data)
            return _object

        } catch {
            return nil
        }
    }
}

var jsonString = "[\n" +
    "{\n" +
    "\"id\":72,\n" +
    "\"name\":\"Batata Cremosa\",\n" +
    "},\n" +
    "{\n" +
    "\"id\":183,\n" +
    "\"name\":\"Caldeirada de Peixes\",\n" +
    "},\n" +
    "{\n" +
    "\"id\":76,\n" +
    "\"name\":\"Batata com Cebola e Ervas\",\n" +
    "},\n" +
    "{\n" +
    "\"id\":56,\n" +
    "\"name\":\"Arroz de forma\",\n" +
"}]"

let businesses = jsonString.parse(to: [Business].self)

Answer 8

对于iOS 10和Swift 3 ，使用Alamofire和Gloss ：

Alamofire.request("http://localhost:8080/category/en").responseJSON { response in

if let data = response.data {

    if let categories = [Category].from(data: response.data) {

        self.categories = categories

        self.categoryCollectionView.reloadData()
    } else {

        print("Casting error")
    }
  } else {

    print("Data is null")
  }
}

这是类别类

import Gloss

struct Category: Decodable {

    let categoryId: Int?
    let name: String?
    let image: String?

    init?(json: JSON) {
        self.categoryId = "categoryId" <~~ json
        self.name = "name" <~~ json
        self.image = "image" <~~ json
    }
}

IMO，这是迄今为止最优雅的解决方案。

Answer 9

let jsonString = "{\"id\":123,\"Name\":\"Munish\"}"

将字符串转换为 NSData

 var data: NSData =jsonString.dataUsingEncoding(NSUTF8StringEncoding)!

 var error: NSError?

将 NSData 转换为 AnyObject

var jsonObject: AnyObject? = NSJSONSerialization.JSONObjectWithData(data,     options: NSJSONReadingOptions.allZeros, error: &error)

println("Error: \\(error)")

let id = (jsonObject as! NSDictionary)["id"] as! Int

let name = (jsonObject as! NSDictionary)["name"] as! String

println("Id: \\(id)")

println("Name: \\(name)")

Answer 10

我喜欢 RDC 的响应，但为什么将返回的 JSON 限制为只有顶级数组？ 我需要在顶层允许字典，所以我修改了它：

extension String
{
    var parseJSONString: AnyObject?
    {
        let data = self.dataUsingEncoding(NSUTF8StringEncoding, allowLossyConversion: false)

        if let jsonData = data
        {
            // Will return an object or nil if JSON decoding fails
            do
            {
                let message = try NSJSONSerialization.JSONObjectWithData(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
        }
    }

Answer 11

SWIFT4 - 将 JSON 字符串解码为 Struct 的简单而优雅的方法。

第一步- 使用 .utf8 编码将字符串编码为数据。

比将您的数据解码为 YourDataStruct。

struct YourDataStruct: Codable {

let type, id: String

init(_ json: String, using encoding: String.Encoding = .utf8) throws {
    guard let data = json.data(using: encoding) else {
        throw NSError(domain: "JSONDecoding", code: 0, userInfo: nil)
    }
    try self.init(data: data)
}

init(data: Data) throws {
    self = try JSONDecoder().decode(YourDataStruct.self, from: data)
}                                                                      
}

do { let successResponse = try WSDeleteDialogsResponse(response) }
} catch {}

Answer 12

这可能是帮助某人。 类似的例子。

这是我们用于绑定数据的Codable类。 您可以使用SwiftyJsonAccelerator轻松创建此类

 class ModelPushNotificationFilesFile: Codable {

  enum CodingKeys: String, CodingKey {
    case url
    case id
    case fileExtension = "file_extension"
    case name
  }

  var url: String?
  var id: Int?
  var fileExtension: String?
  var name: String?

  init (url: String?, id: Int?, fileExtension: String?, name: String?) {
    self.url = url
    self.id = id
    self.fileExtension = fileExtension
    self.name = name
  }

  required init(from decoder: Decoder) throws {
    let container = try decoder.container(keyedBy: CodingKeys.self)
    url = try container.decodeIfPresent(String.self, forKey: .url)
    id = try container.decodeIfPresent(Int.self, forKey: .id)
    fileExtension = try container.decodeIfPresent(String.self, forKey: .fileExtension)
    name = try container.decodeIfPresent(String.self, forKey: .name)
  }

}

这是 Json 字符串

    let jsonString = "[{\"name\":\"\",\"file_extension\":\"\",\"id\":10684,\"url\":\"https:\\/\\/homepages.cae.wisc.edu\\/~ece533\\/images\\/tulips.png\"},

{\"name\":\"\",\"file_extension\":\"\",\"id\":10684,\"url\":\"https:\\/\\/homepages.cae.wisc.edu\\/~ece533\\/images\\/arctichare.png\"},

{\"name\":\"\",\"file_extension\":\"\",\"id\":10684,\"url\":\"https:\\/\\/homepages.cae.wisc.edu\\/~ece533\\/images\\/serrano.png\"},

{\"name\":\"\",\"file_extension\":\"\",\"id\":10684,\"url\":\"https:\\/\\/homepages.cae.wisc.edu\\/~ece533\\/images\\/peppers.png\"},

{\"name\":\"\",\"file_extension\":\"\",\"id\":10684,\"url\":\"https:\\/\\/homepages.cae.wisc.edu\\/~ece533\\/images\\/pool.png\"}]"

这里我们转换为 swift 对象。

   let jsonData = Data(jsonString.utf8)

        let decoder = JSONDecoder()

        do {
            let fileArray = try decoder.decode([ModelPushNotificationFilesFile].self, from: jsonData)
            print(fileArray)
            print(fileArray[0].url)
        } catch {
            print(error.localizedDescription)
        }

Answer 13

您可以使用swift.quicktype.io将JSON转换为struct或class 。 甚至你可以提到 swift 的版本来生成代码。

示例 JSON：

{
  "message": "Hello, World!"
}

生成的代码：

import Foundation

typealias Sample = OtherSample

struct OtherSample: Codable {
    let message: String
}

// Serialization extensions

extension OtherSample {
    static func from(json: String, using encoding: String.Encoding = .utf8) -> OtherSample? {
        guard let data = json.data(using: encoding) else { return nil }
        return OtherSample.from(data: data)
    }

    static func from(data: Data) -> OtherSample? {
        let decoder = JSONDecoder()
        return try? decoder.decode(OtherSample.self, from: data)
    }

    var jsonData: Data? {
        let encoder = JSONEncoder()
        return try? encoder.encode(self)
    }

    var jsonString: String? {
        guard let data = self.jsonData else { return nil }
        return String(data: data, encoding: .utf8)
    }
}

extension OtherSample {
    enum CodingKeys: String, CodingKey {
        case message
    }
}

Answer 14

使用SwiftyJSON库，你可以让它像

if let path : String = Bundle.main.path(forResource: "tiles", ofType: "json") {
    if let data = NSData(contentsOfFile: path) {
        let optData = try? JSON(data: data as Data)
        guard let json = optData else {
            return
        }
        for (_, object) in json {
            let name = object["name"].stringValue
            print(name)
        }
    }
} 

Answer 15

这里有一个示例，可让您更简单、更轻松地完成工作。 我的数据库中的字符串数据是一个 JSON 文件，如下所示：

[{"stype":"noun","sdsc":"careless disregard for consequences","swds":"disregard, freedom, impulse, licentiousness, recklessness, spontaneity, thoughtlessness, uninhibitedness, unrestraint, wantonness, wildness","anwds":"restraint, self-restraint"},{"stype":"verb","sdsc":"leave behind, relinquish","swds":"abdicate, back out, bail out, bow out, chicken out, cop out, cut loose, desert, discard, discontinue, ditch, drop, drop out, duck, dump, dust, flake out, fly the coop, give up the ship, kiss goodbye, leave, leg it, let go, opt out, pull out, quit, run out on, screw, ship out, stop, storm out, surrender, take a powder, take a walk, throw over, vacate, walk out on, wash hands of, withdraw, yield","anwds":"adopt, advance, allow, assert, begin, cherish, come, continue, defend, favor, go, hold, keep, maintain, persevere, pursue, remain, retain, start, stay, support, uphold"},{"stype":"verb","sdsc":"leave in troubled state","swds":"back out, desert, disown, forsake, jilt, leave, leave behind, quit, reject, renounce, throw over, walk out on","anwds":"adopt, allow, approve, assert, cherish, come, continue, defend, favor, keep, pursue, retain, stay, support, uphold"}]

要加载此 JSON 字符串数据，请按照以下简单步骤操作。 首先，为我的 MoreData 对象创建一个类，如下所示：

class  MoreData {
public private(set) var stype : String
public private(set) var sdsc : String
public private(set) var swds : String
public private(set) var anwds : String

init( stype : String, sdsc : String, swds : String, anwds : String) {

    self.stype = stype
    self.sdsc = sdsc
    self.swds = swds
    self.anwds = anwds
}}

其次，为我的 JSON 字符串创建我的字符串扩展，如下所示：

extension  String {
func toJSON() -> Any? {
    guard let data = self.data(using: .utf8, allowLossyConversion: false) else { return nil }
    return try? JSONSerialization.jsonObject(with: data, options: .mutableContainers)
}}

第三，创建我的服务类来处理我的字符串数据，如下所示：

class Services {
static let instance: Services = Services()

func loadMoreDataByString(byString: String) -> [MoreData]{
    var  myVariable = [MoreData]()

    guard let ListOf = byString.toJSON() as? [[String: AnyObject]] else { return  [] }

    for object in ListOf {
        let stype  = object["stype"] as? String ?? ""
        let sdsc  = object["sdsc"] as? String ?? ""
         let swds  = object["swds"] as? String ?? ""
        let anwds  = object["anwds"] as? String ?? ""

        let myMoreData = MoreData(stype : stype, sdsc : sdsc, swds : swds, anwds : anwds)
        myVariable.append(myMoreData)
    }
    return myVariable
}}

最后，从视图控制器调用此函数以在表视图中加载数据，如下所示：

    func handlingJsonStringData(){
    moreData.removeAll(keepingCapacity: false)
    moreData =  Services.instance.loadMoreDataByString(byString: jsonString)
    print(self.moreData.count)
    tableView.reloadData()
}

Answer 16

使用 swiftyJson swiftyJson

platform :ios, '8.0'
use_frameworks!

target 'MyApp' do
pod 'SwiftyJSON', '~> 4.0'
end

用法

import SwiftyJSON

let json = JSON(jsonObject)

let id = json["Id"].intValue
let name = json["Name"].stringValue
let lat = json["Latitude"].stringValue
let long = json["Longitude"].stringValue
let address = json["Address"].stringValue
            
print(id)
print(name)
print(lat)
print(long)
print(address)

Answer 17

将 json 包装在多行字符串文字中，并尝试像这样解析：

import Foundation

// MARK: - DemoJSON
struct DemoJSON: Codable {
    let menu: Menu
}

// MARK: - Menu
struct Menu: Codable {
    let id, value: String
    let popup: Popup
}

// MARK: - Popup
struct Popup: Codable {
    let menuitem: [Menuitem]
}

// MARK: - Menuitem
struct Menuitem: Codable {
    let value, onclick: String
}



/// Make Network Request
typealias RequestCompletionHandler<T: Decodable> = (_ value: T?, _ error: Error?) -> Void


func callAPI<T: Decodable>(completionHandler: RequestCompletionHandler<T>) {
    let data = """
{"menu": {
  "id": "file",
  "value": "File",
  "popup": {
    "menuitem": [
      {"value": "New", "onclick": "CreateNewDoc()"},
      {"value": "Open", "onclick": "OpenDoc()"},
      {"value": "Close", "onclick": "CloseDoc()"}
    ]
  }
}}
""".data(using: .utf8)!
    do {
        let value = try JSONDecoder().decode(T.self, from: data)
        completionHandler(value, nil)
    } catch {
        completionHandler(nil, error)
    }
}

callAPI { (model: DemoJSON?, error) in
    if let demoModel = model {
        print("Success: \(demoModel)")
    } else if let error = error {
        print("Error: \(error)")
    }
}