在Swift 2中将JSON转换为数组

Question

我需要构建Arrays一个Grouped UITableView,每个表格单元格中都有一个Title和Detail行.我从服务器获得了json输出,使其形状正确,以迭代UITableViewDataSource方法.但是,将这些UITableView函数引用的可读数组转换为最简单的方法是什么？

标题数组用于组标题,因此它只是一维数组.我可以迭代一下.标题和细节数组都是二维的.我无法弄清楚如何在Swift中做到这一点.

"headings":["Tuesday, August 16, 2016","Wednesday, August 17, 2016","Thursday, August 18, 2016","Friday, August 19, 2016","Saturday, August 20, 2016","Sunday, August 21, 2016","Monday, August 22, 2016","Tuesday, August 23, 2016","Wednesday, August 24, 2016","Thursday, August 25, 2016","Friday, August 26, 2016","Saturday, August 27, 2016","Sunday, August 28, 2016","Monday, August 29, 2016","Tuesday, August 30, 2016","Wednesday, August 31, 2016","Thursday, September 1, 2016","Friday, September 2, 2016","Saturday, September 3, 2016","Sunday, September 4, 2016","Monday, September 5, 2016","Tuesday, September 6, 2016","Wednesday, September 7, 2016","Thursday, September 8, 2016","Friday, September 9, 2016","Saturday, September 10, 2016","Sunday, September 11, 2016","Monday, September 12, 2016","Tuesday, September 13, 2016","Wednesday, September 14, 2016","Thursday, September 15, 2016","Friday, September 16, 2016"],

"titles":[["Joe Johnson"],["Joe Johnson"],["Sandy Primmell","Joe Johnson"],["Joe Johnson"],["Joe Johnson"],["Joe Johnson"],["Joe Johnson"],["Sandy Primmell","Joe Johnson"],["Joe Johnson","Joe Johnson"],["Sandy Primmell","Joe Johnson"],["Mark Greene","Joe Johnson"],["Joe Johnson"],["Joe Johnson"],["Joe Johnson"],["Joe Johnson"],["Sandy Primmell","Joe Johnson"],["Joe Johnson"],["Sandy Primmell","Joe Johnson"],["Mark Greene","Joe Johnson"],["Joe Johnson"],["Joe Johnson"],["Joe Johnson"],["Joe Johnson"],["Joe Johnson"],["Joe Johnson"],["Joe Johnson"],["Joe Johnson"],["Joe Johnson"],["Joe Johnson"],["Joe Johnson"],["Joe Johnson"],["Joe Johnson"]],

"details":[["OFF"],["OFF"],["Gregory","OFF"],["Gregory"],["OFF"],["OFF"],["OFF"],["Weekday Rounders","OFF"],["Weekday Rounders","Night Owls"],["Gregory","OFF"],["Gregory","OFF"],["OFF"],["OFF"],["OFF"],["Gregory"],["Gregory","OFF"],["Gregory"],["Gregory","OFF"],["Gregory","OFF"],["OFF"],["OFF"],["OFF"],["OFF"],["OFF"],["OFF"],["OFF"],["OFF"],["OFF"],["OFF"],["OFF"],["OFF"],["OFF"]]

UPDATE

这是抓取数据的Alamofire异步函数:

manager.request(.POST, getRouter(), parameters:["dev": 1, "app_action": "schedule", "type":getScheduleType(), "days_off":getScheduleDaysOff(), "period":getSchedulePeriod(), "begin_date":getScheduleBeginDate(), "end_date":getScheduleEndDate()])
        .responseString {response in
            print(response)
            var json = JSON(response.result.value!);
// what I'm missing
   }

Answer 1

或者，您可以使用Argo或SwiftyJSON等 JSON 解析库，它们是为了简化 JSON 解析而创建的。它们都经过了良好的测试，并且会为您处理边缘情况，例如 JSON 响应中缺少参数等。

使用 Argo 的示例：

假设 JSON 响应具有这种格式（来自Twitter API）

{
  "users": [
    {
      "id": 2960784075,
      "id_str": "2960784075",
      ...
    }
}

1- 创建一个 Swift 类来表示响应

请注意，这Response是一个包含数组的类，User该数组是此处未显示的另一个类，但您明白了。

struct Response: Decodable {
    let users: [User]
    let next_cursor_str: String

    static func decode(j: JSON) -> Decoded<Response> {
        return curry(Response.init)
            <^> j <|| "users"
            <*> j <| "next_cursor_str"
    }
}

2-解析 JSON

//Convert json String to foundation object
let json: AnyObject? = try? NSJSONSerialization.JSONObjectWithData(data, options: [])

//Check for nil    
if let j: AnyObject = json {
  //Map the foundation object to Response object
  let response: Response? = decode(j)
}

使用 Swifty 的示例

正如官方文档中所解释的：

1-将 JSON 字符串转换为 SwiftyJSON 对象

if let dataFromString = jsonString.dataUsingEncoding(NSUTF8StringEncoding, allowLossyConversion: false) {
    let json = JSON(data: dataFromString)
}

2-访问特定元素

如果数据是数组，则使用索引

//Getting a double from a JSON Array
let name = json[0].double

如果数据是字典，则使用键

//Getting a string from a JSON Dictionary
let name = json["name"].stringValue

2'- 循环元素

大批

//If json is .Array
//The `index` is 0..<json.count's string value
for (index,subJson):(String, JSON) in json {
    //Do something you want
}

字典

//If json is .Dictionary
for (key,subJson):(String, JSON) in json {
   //Do something you want
}