在Swift中将JSON字符串转换为Object的简单方法

82

我一直在寻找几天，以便在Swift中将相当简单的JSON字符串转换为对象类型，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)

        }
}

我有一个快速的结构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
  }
  ...
]

如果有人通过这个指导我会很高兴。

谢谢。

ios  json  swift  iphone 
— 哈桑·尼扎马尼（Hasan Nizamani）
source

Answers:

56

这里有一些技巧，以简单的例子开始。

考虑您具有以下JSON数组字符串（类似于您的）：

 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]

摆脱null更改为：

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

另请参阅Coalescing Operator（aka的参考??）

希望它能帮助您解决问题，

— 马克西姆·肖斯汀
source
太棒了！像魅力一样工作。谢谢！一件很小的事情，如果JSON中的元素为null，就会出现错误。就像在：b.name = json [“ name”]中那样作为AnyObject！as String如果name为null，如何添加条件使其可以为null？
— Hasan Nizamani 2014年
在强制转换为String之前需要强制转换为AnyObject？
— Bateramos 2015年
@Bateramos没有。您可以AnyObject通过键获得可选的选项，只是在向下转换到String之前确定不是nil。出于这个原因，我可以使用!或在场所中使用?capsualte??
— Maxim Shoustin 15-2-4
我猜想，如果您创建对象var b:Business = Business()外循环，那么它可能在每个列表元素中显示相同的数据。
— Patriks
50

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)
    }
}

用法示例：

 let dict = myString.toJSON() as? [String:AnyObject] // can be any type here
— FlowUI。SimpleUITesting.com
source
2
未来的注脚：不是do-catch，try?这里可以使用，这将给予同样的结果在返回nil catch。
— Okhan Okbay
...以及这种转换后究竟如何访问一个参数？
— Starwave
对于其他感兴趣的人：让jsonObjectAsNSDictionary = responseString？.toJSON（）为！[String：AnyObject] print（jsonObjectAsNSDictionary [“ permissions”]！[“ canaddeditowncomment”]）
— Starwave
1
请更正快速语法... 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) } }
— Yasir Ali，
27

由于简单的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";
}
*/
— 通行证
source
15

Swift 4更优雅地解析JSON。只需按照此简化示例为您的结构采用可编码协议：

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

要解析JSON数组，您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，请查看此出色的指南。

— Noyer282
source
这样做的缺点是必须跟踪该结构，如果您拥有约30个或更多参数，则管理将变得很麻烦。
— Starwave
@Starwave：不确定是否可以解决您的问题，但是请注意，您只需要在用于解码数据的结构中包括您关心的字段即可。例如，如果JSON的格式为[{“ id”：1，“ name：” A“，” location“：” Warsaw“}，]，您仍然可以使用相同的Business结构对其进行解码。location字段将只是忽略。
— Noyer282
血腥的地狱，我没想到……但是，这里的parsedData是什么？NSDictionary？
— Starwave
@Starwave：这是一个结构数组，就像原始问题一样。
— Noyer282 '18
9

对于Swift 4

我使用了@ Passkit的逻辑，但是我必须按照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使用的方式

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])")

所有功劳归原始用户所有，我刚刚更新为最新的快速版本

— 迅捷男孩
source
7

我编写了一个库，使在Swift中轻松处理json数据和反序列化变得很容易。您可以在这里获取它：https : //github.com/isair/JSONHelper

编辑：我更新了我的库，您现在可以使用以下方法：

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，而要使用.response获取响应对象。然后将代码更改为：

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. 
        }
    }
}

您需要制作一个如下的商务舱：

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存储库中找到完整的文档。玩得开心！

— 伊塞尔
source
谢谢！但是，我一直在寻找一种自定义方式来执行此操作，因为这是我正在从事的大学项目，而我的主管如果看到我正在使用API​​，不会很高兴。
— 哈桑·尼扎马尼
将在我的私人项目上使用您的库。谢谢！！
— Hasan Nizamani 2014年
很高兴我能够以某种方式提供帮助。祝您项目顺利！^^
— isair 2014年
6

对于Swift 4，我使用Codable协议编写了此扩展：

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)
— 丹尼·纳瓦兹（DannyNarváez）
source
4

对于iOS 10＆Swift 3，使用Alamofire＆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。

— 永恒的
source
2 
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)")
— 穆尼什·卡普尔
source
2

我喜欢RDC的响应，但是为什么限制返回的JSON仅在顶层具有数组？我需要在顶层允许使用字典，因此我对其进行了修改：

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
        }
    }
— MR_22
source
2

SWIFT4-将JSON字符串解码为Struct的简便方法。

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 {}
— 维蒂亚·舒拉波夫（Vitya Shurapov）
source
1

您可以使用swift.quicktype.io转换JSON为struct或class。甚至您都可以提到Swift版本来生成代码。

JSON示例： 

{
  "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
    }
}
— Rugmangathan
source
1

使用SwiftyJSON库，您可以像

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)
        }
    }
}
— 穆罕默德·萨利赫（Mohamed Saleh）
source
0

这是让您使事情变得简单和容易的示例。我在数据库中的String数据是一个如下所示的JSON文件：

[{"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，请遵循以下简单步骤。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 String创建String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My Srevices类来处理我的String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，以将数据加载到表视图中，如下所示：

    func handlingJsonStringData(){
    moreData.removeAll(keepingCapacity: false)
    moreData =  Services.instance.loadMoreDataByString(byString: jsonString)
    print(self.moreData.count)
    tableView.reloadData()
}
— 哈迪巴利
source
0

可能对某人有所帮助。类似的例子。

这是我们的Codable绑定数据类。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 String

    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\"}]"

在这里，我们将转换为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)
        }
— Shourob Datta
source
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