快速创建 JSON

Question

我需要像这样创建 JSON：

Order = {   type_id:'1',model_id:'1',

   transfer:{
     startDate:'10/04/2015 12:45',
     endDate:'10/04/2015 16:00',
     startPoint:'Ул. Момышулы, 45',
     endPoint:'Аэропорт Астаны'
   },
   hourly:{
     startDate:'10/04/2015',
     endDate:'11/04/2015',
     startPoint:'ЖД Вокзал',
     endPoint:'',
     undefined_time:'1'
   },
   custom:{
     startDate:'12/04/2015',
     endDate:'12/04/2015',
     startPoint:'Астана',
     endPoint:'Павлодар',
     customPrice:'50 000'
   },
    commentText:'',
    device_type:'ios'
};

问题是我无法创建有效的 JSON。 以下是我创建对象的方式：

let jsonObject: [AnyObject]  = [
        ["type_id": singleStructDataOfCar.typeID, "model_id": singleStructDataOfCar.modelID, "transfer": savedDataTransfer, "hourly": savedDataHourly, "custom": savedDataReis, "device_type":"ios"]
    ]

savedData 是字典：

let savedData: NSDictionary = ["ServiceDataStartDate": singleStructdata.startofWork, 
"ServiceDataAddressOfReq": singleStructdata.addressOfRequest, 
"ServiceDataAddressOfDel": singleStructdata.addressOfDelivery, 
"ServiceDataDetailedText": singleStructdata.detailedText, "ServiceDataPrice": singleStructdata.priceProposed]

当我只使用字符串创建我的 JSON 对象时，一切正常。但是，当我包含字典 @​​987654326@ 时，返回 false。如何创建有效的字典？

Answer 1

斯威夫特 5 - 21 年 6 月 30 日

采用 Codable 协议（类似于其他编程语言中的接口）

struct ConfigRequestBody: Codable {
var systemid: String
var password: String
var request: String = "getconfig"

init(systemID: String, password: String){
    self.systemid = systemID
    self.password = password
}

}

创建要转换为 JSON 的结构/类的实例：

let requestBody = ConfigRequestBody(systemID: systemID, password: password)

使用JSONEncoder 将对象编码为 JSON。这里我打印字符串表示，以便您查看结果：

    let encoder = JSONEncoder()
    encoder.outputFormatting = .prettyPrinted
    do {
        let result = try encoder.encode(requestBody)
        // RESULT IS NOW JSON-LIKE DATA OBJECT
        if let jsonString = String(data: result, encoding: .utf8){
            // JSON STRING
            print("JSON \(jsonString)")
        }
    } catch {
        print("Your parsing sucks \(error)")
        return nil
    }
}

Answer 2

对于 Swift 3.0，截至 2016 年 12 月，这对我来说是这样的：

let jsonObject: NSMutableDictionary = NSMutableDictionary()

jsonObject.setValue(value1, forKey: "b")
jsonObject.setValue(value2, forKey: "p")
jsonObject.setValue(value3, forKey: "o")
jsonObject.setValue(value4, forKey: "s")
jsonObject.setValue(value5, forKey: "r")

let jsonData: NSData

do {
    jsonData = try JSONSerialization.data(withJSONObject: jsonObject, options: JSONSerialization.WritingOptions()) as NSData
    let jsonString = NSString(data: jsonData as Data, encoding: String.Encoding.utf8.rawValue) as! String
    print("json string = \(jsonString)")                                    

} catch _ {
    print ("JSON Failure")
}

EDIT 2018：我现在使用 SwiftyJSON 库来节省时间，让我的开发生活更轻松、更好。在 Swift 中原生处理 JSON 是一种不必要的头痛和痛苦，而且浪费太多时间，并且创建的代码难以读写，因此容易出错。

Answer 3

• Swift 4.1，2018 年 4 月

这是一种更通用的方法，可用于通过使用字典中的值来创建 JSON 字符串：

struct JSONStringEncoder {
    /**
     Encodes a dictionary into a JSON string.
     - parameter dictionary: Dictionary to use to encode JSON string.
     - returns: A JSON string. `nil`, when encoding failed.
     */
    func encode(_ dictionary: [String: Any]) -> String? {
        guard JSONSerialization.isValidJSONObject(dictionary) else {
            assertionFailure("Invalid json object received.")
            return nil
        }

        let jsonObject: NSMutableDictionary = NSMutableDictionary()
        let jsonData: Data

        dictionary.forEach { (arg) in
            jsonObject.setValue(arg.value, forKey: arg.key)
        }

        do {
            jsonData = try JSONSerialization.data(withJSONObject: jsonObject, options: .prettyPrinted)
        } catch {
            assertionFailure("JSON data creation failed with error: \(error).")
            return nil
        }

        guard let jsonString = String.init(data: jsonData, encoding: String.Encoding.utf8) else {
            assertionFailure("JSON string creation failed.")
            return nil
        }

        print("JSON string: \(jsonString)")
        return jsonString
    }
}

如何使用：

let exampleDict: [String: Any] = [
        "Key1" : "stringValue",         // type: String
        "Key2" : boolValue,             // type: Bool
        "Key3" : intValue,              // type: Int
        "Key4" : customTypeInstance,    // type: e.g. struct Person: Codable {...}
        "Key5" : customClassInstance,   // type: e.g. class Human: NSObject, NSCoding {...}
        // ... 
    ]

    if let jsonString = JSONStringEncoder().encode(exampleDict) {
        // Successfully created JSON string.
        // ... 
    } else {
        // Failed creating JSON string.
        // ...
    }

注意：如果您将自定义类型（结构）的实例添加到字典中，请确保您的类型符合 Codable 协议，如果您将自定义类的对象添加到字典中，请确保您的类继承自NSObject 并符合NSCoding 协议。

Answer 4

一个问题是这段代码不是Dictionary类型的。

let jsonObject: [Any]  = [
    [
         "type_id": singleStructDataOfCar.typeID,
         "model_id": singleStructDataOfCar.modelID, 
         "transfer": savedDataTransfer, 
         "hourly": savedDataHourly, 
         "custom": savedDataReis, 
         "device_type":"iOS"
    ]
]

上面是AnyObject 的Array，其中包含[String: AnyObject] 类型的Dictionary。

尝试这样的方式来匹配您在上面提供的 JSON：

let savedData = ["Something": 1]

let jsonObject: [String: Any] = [ 
    "type_id": 1,
    "model_id": 1,
    "transfer": [
        "startDate": "10/04/2015 12:45",
        "endDate": "10/04/2015 16:00"
    ],
    "custom": savedData
]

let valid = JSONSerialization.isValidJSONObject(jsonObject) // true

Answer 5

创建 JSON 字符串：

let para:NSMutableDictionary = NSMutableDictionary()
para.setValue("bidder", forKey: "username")
para.setValue("day303", forKey: "password")
para.setValue("authetication", forKey: "action")
let jsonData = try! NSJSONSerialization.dataWithJSONObject(para, options: NSJSONWritingOptions.allZeros)
let jsonString = NSString(data: jsonData, encoding: NSUTF8StringEncoding) as! String
print(jsonString)

Answer 6

查看https://github.com/peheje/JsonSerializerSwift

用例：

//Arrange your model classes
class Object {
  var id: Int = 182371823
  }
class Animal: Object {
  var weight: Double = 2.5
  var age: Int = 2
  var name: String? = "An animal"
  }
class Cat: Animal {
  var fur: Bool = true
}

let m = Cat()

//Act
let json = JSONSerializer.toJson(m)

//Assert
let expected = "{\"fur\": true, \"weight\": 2.5, \"age\": 2, \"name\": \"An animal\", \"id\": 182371823}"
stringCompareHelper(json, expected) //returns true

目前支持标准类型、可选标准类型、数组、可空标准类型数组、自定义类数组、继承、自定义对象组合。

Answer 7

这对我有用... Swift 2

static func checkUsernameAndPassword(username: String, password: String) -> String?{
    let para:NSMutableDictionary = NSMutableDictionary()
        para.setValue("demo", forKey: "username")
        para.setValue("demo", forKey: "password")
       // let jsonError: NSError?
    let jsonData: NSData
    do{
        jsonData = try NSJSONSerialization.dataWithJSONObject(para, options: NSJSONWritingOptions())
        let jsonString = NSString(data: jsonData, encoding: NSUTF8StringEncoding) as! String
        print("json string = \(jsonString)")
        return jsonString

    } catch _ {
        print ("UH OOO")
        return nil
    }
}