我对你的问题的回答如下

let dict = ["0": "ArrayObjectOne", "1": "ArrayObjecttwo", "2": "ArrayObjectThree"]

var error : NSError?

let jsonData = try! NSJSONSerialization.dataWithJSONObject(dict, options: NSJSONWritingOptions.PrettyPrinted)

let jsonString = NSString(data: jsonData, encoding: String.Encoding.utf8.rawValue)! as String

print(jsonString)

答案是

{
  "0" : "ArrayObjectOne",
  "1" : "ArrayObjecttwo",
  "2" : "ArrayObjectThree"
}

斯威夫特3.0

在Swift 3中，NSJSONSerialization的名字和它的方法已经改变了，根据Swift API设计指南。

let dic = ["2": "B", "1": "A", "3": "C"]

do {
    let jsonData = try JSONSerialization.data(withJSONObject: dic, options: .prettyPrinted)
    // here "jsonData" is the dictionary encoded in JSON data

    let decoded = try JSONSerialization.jsonObject(with: jsonData, options: [])
    // here "decoded" is of type `Any`, decoded from JSON data

    // you can now cast it with the right type        
    if let dictFromJSON = decoded as? [String:String] {
        // use dictFromJSON
    }
} catch {
    print(error.localizedDescription)
}

快2.倍

do {
    let jsonData = try NSJSONSerialization.dataWithJSONObject(dic, options: NSJSONWritingOptions.PrettyPrinted)
    // here "jsonData" is the dictionary encoded in JSON data

    let decoded = try NSJSONSerialization.JSONObjectWithData(jsonData, options: [])
    // here "decoded" is of type `AnyObject`, decoded from JSON data

    // you can now cast it with the right type 
    if let dictFromJSON = decoded as? [String:String] {
        // use dictFromJSON
    }
} catch let error as NSError {
    print(error)
}

斯威夫特1

var error: NSError?
if let jsonData = NSJSONSerialization.dataWithJSONObject(dic, options: NSJSONWritingOptions.PrettyPrinted, error: &error) {
    if error != nil {
        println(error)
    } else {
        // here "jsonData" is the dictionary encoded in JSON data
    }
}

if let decoded = NSJSONSerialization.JSONObjectWithData(jsonData, options: nil, error: &error) as? [String:String] {
    if error != nil {
        println(error)
    } else {
        // here "decoded" is the dictionary decoded from JSON data
    }
}

有时为了调试目的，有必要打印出服务器的响应。下面是我使用的函数:

extension Dictionary {

    var json: String {
        let invalidJson = "Not a valid JSON"
        do {
            let jsonData = try JSONSerialization.data(withJSONObject: self, options: .prettyPrinted)
            return String(bytes: jsonData, encoding: String.Encoding.utf8) ?? invalidJson
        } catch {
            return invalidJson
        }
    }

    func printJson() {
        print(json)
    }

}

使用示例:

(lldb) po dictionary.printJson()
{
  "InviteId" : 2,
  "EventId" : 13591,
  "Messages" : [
    {
      "SenderUserId" : 9514,
      "MessageText" : "test",
      "RecipientUserId" : 9470
    },
    {
      "SenderUserId" : 9514,
      "MessageText" : "test",
      "RecipientUserId" : 9470
    }
  ],
  "TargetUserId" : 9470,
  "InvitedUsers" : [
    9470
  ],
  "InvitingUserId" : 9514,
  "WillGo" : true,
  "DateCreated" : "2016-08-24 14:01:08 +00:00"
}

你问题的答案如下:

斯威夫特2.1

     do {
          if let postData : NSData = try NSJSONSerialization.dataWithJSONObject(dictDataToBeConverted, options: NSJSONWritingOptions.PrettyPrinted){

          let json = NSString(data: postData, encoding: NSUTF8StringEncoding)! as String
          print(json)}

        }
        catch {
           print(error)
        }

使用lldb

(lldb) p JSONSerialization.data(withJSONObject: notification.request.content.userInfo, options: [])
(Data) $R16 = 375 bytes
(lldb) p String(data: $R16!, encoding: .utf8)!
(String) $R18 = "{\"aps\": \"some_text\"}"

//or
p String(data: JSONSerialization.data(withJSONObject: notification.request.content.userInfo, options: [])!, encoding: .utf8)!
(String) $R4 = "{\"aps\": \"some_text\"}"

这对我来说很管用:

import SwiftyJSON

extension JSON {
    
    mutating func appendIfKeyValuePair(key: String, value: Any){
        if var dict = self.dictionaryObject {
            dict[key] = value
            self = JSON(dict)
        }
    }
}

用法:

var data: JSON = []

data.appendIfKeyValuePair(key: "myKey", value: "myValue")