在swift中创建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:'??. ????????, 45',
     endPoint:'???????? ??????'
   },
   hourly:{
     startDate:'10/04/2015',
     endDate:'11/04/2015',
     startPoint:'?? ??????',
     endPoint:'',
     undefined_time:'1'
   },
   custom:{
     startDate:'12/04/2015',
     endDate:'12/04/2015',
     startPoint:'??????',
     endPoint:'????????',
     customPrice:'50 000'
   },
    commentText:'',
    device_type:'ios'
};

问题是我无法创建有效的JSON.这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词典在哪里:

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.但是当我包含词典NSJSONSerialization.isValidJSONObject(value)返回时false.如何创建有效的字典？

Answer 1

一个问题是此代码不是类型Dictionary.

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

以上是一个Array的AnyObject与Dictionary类型的[String: AnyObject]在其内部.

尝试这样的方法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 2

对于Swift 3.0,截至2016年12月,这是它对我有用的方式:

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")
}

编辑2018:我转向SwiftyJSON以节省时间并使我的开发生活更轻松,更好.在Swift中本地处理JSON是一种不必要的头痛和痛苦,加上浪费了太多时间,并且创建了难以读写的代码,因此容易出现大量错误.

Answer 3

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 4

\xe2\x80\xa2 Swift 4.1，2018 年 4 月

\n\n

以下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：

\n\n

struct JSONStringEncoder {\n    /**\n     Encodes a dictionary into a JSON string.\n     - parameter dictionary: Dictionary to use to encode JSON string.\n     - returns: A JSON string. `nil`, when encoding failed.\n     */\n    func encode(_ dictionary: [String: Any]) -> String? {\n        guard JSONSerialization.isValidJSONObject(dictionary) else {\n            assertionFailure("Invalid json object received.")\n            return nil\n        }\n\n        let jsonObject: NSMutableDictionary = NSMutableDictionary()\n        let jsonData: Data\n\n        dictionary.forEach { (arg) in\n            jsonObject.setValue(arg.value, forKey: arg.key)\n        }\n\n        do {\n            jsonData = try JSONSerialization.data(withJSONObject: jsonObject, options: .prettyPrinted)\n        } catch {\n            assertionFailure("JSON data creation failed with error: \\(error).")\n            return nil\n        }\n\n        guard let jsonString = String.init(data: jsonData, encoding: String.Encoding.utf8) else {\n            assertionFailure("JSON string creation failed.")\n            return nil\n        }\n\n        print("JSON string: \\(jsonString)")\n        return jsonString\n    }\n}\n

\n\n

如何使用它：

\n\n

let exampleDict: [String: Any] = [\n        "Key1" : "stringValue",         // type: String\n        "Key2" : boolValue,             // type: Bool\n        "Key3" : intValue,              // type: Int\n        "Key4" : customTypeInstance,    // type: e.g. struct Person: Codable {...}\n        "Key5" : customClassInstance,   // type: e.g. class Human: NSObject, NSCoding {...}\n        // ... \n    ]\n\n    if let jsonString = JSONStringEncoder().encode(exampleDict) {\n        // Successfully created JSON string.\n        // ... \n    } else {\n        // Failed creating JSON string.\n        // ...\n    }\n

\n\n

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

\n

Answer 5

Swift 5 - 6/30/21

Adopt the Codable protocol (similar to interface in other programming languages)

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

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

Create instance of struct/class that you want to turn into JSON:

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

Encode the object into JSON using a JSONEncoder. Here I print the string representation so you can see the result:

    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
    }
}