在Swift中将JSON字符串转换为Object的简单而干净的方法

Question

我一直在寻找几天将一个相当简单的JSON字符串转换为Swift中的对象类型,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
  }
  ...
]

如果有人引导我完成这件事,那将是一件令人高兴的事.

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):

 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]

要删除null更改为:

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

参见Coalescing Operator(又名??)参考

希望它能帮助你解决问题,

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)
    }
}

用法示例:

 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

对于Swift 4

我使用@ Passkit的逻辑,但我必须根据Swift 4进行更新

Step.1 为String Class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])")

所有功劳归于原始用户,我刚刚更新了最新的swift版本

Answer 5

Swift 4更优雅地解析JSON.根据此简化示例,只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,请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 6

我写了一个库,它使得使用json数据和反序列化在Swift中变得轻而易举.你可以在这里得到它:https://github.com/isair/JSONHelper

编辑:我更新了我的库,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,使用.response代替使用.responseString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. 
        }
    }
}

你需要创建一个像这样的Business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仓库中找到完整的文档.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& 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 Category 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。