Question

我正在尝试解析并从这个JSON结构中获取值：

["mst_customer": 1, "data": {
    0 = 2;
    1 = 1;
    2 = 1;
    3 = "JAYSON TAMAYO";
    4 = "581-113-113";
    5 = 56;
    6 = on;
    7 = g;
    8 = jayson;
    9 = active;
    "app_access" = on;
    id = 2;
    "mst_customer" = 1;
    name = "Jayson Tamayo";
    status = active;
    territory = 1;
}, "status": OK, "staff_id": 2, "staff_name": Jayson Tamayo]

我使用以下Swift代码来获取值：

    (data: Dictionary<String, AnyObject>, error: String?) -> Void in
                    if error != nil {
                        print(error)
                    } else {

                        if let feed = data["data"] as? NSDictionary ,let entries = data["data"] as? NSArray{
                        for elem: AnyObject in entries{
                              if let staff_name = elem["name"] as? String{
                              print(staff_name)
                              }
                        }
                     }
    }

我试图通过使用键名称或staff_name来获取名称。但我总是没有。

Answer 1

for elem: AnyObject in entries{
    if let songName = elem["name"] as? String{
       print(songName)
    }
}

//replace above code with below code

if let songName : String = entries["name"] as? String{
   print(songName)
}

Answer 2

您想访问staff_name，而不是＆＃34;数据＆＃34;变量...你可以简单地得到它

 if error != nil {
     print(error)
  } else {
      if let name = data["staff_name"] as? String{
          print(name)
      }
  }

解析JSON时的NIL

2 个答案: